River Crossing Problems

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current”

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” There are three relative velocities used. What are they?

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw Bvg wvg

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw = boat’s velocity with respect to water Bvg wvg

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw = boat’s velocity with respect to water the direction of Bvw is called the _________ Bvg wvg

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw = boat’s velocity with respect to water the direction of Bvw is called the heading Bvg wvg

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw = boat’s velocity with respect to water the direction of Bvw is called the heading Heading is also the direction that the boat is pointed Bvg = wvg

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw = boat’s velocity with respect to water the direction of Bvw is called the heading Heading is also the direction that the boat is pointed Bvg = boat’s velocity with respect to the ground or shore wvg

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw = boat’s velocity with respect to water the direction of Bvw is called the heading Heading is also the direction that the boat is pointed Bvg = boat’s velocity with respect to the ground or shore The direction of Bvg is the direction the boat actually moves with respect to a shore or ground observer Wvg =

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw = boat’s velocity with respect to water the direction of Bvw is called the heading Heading is also the direction that the boat is pointed Bvg = boat’s velocity with respect to the ground or shore The direction of Bvg is the direction the boat actually moves with respect to a shore or ground observer Wvg = velocity of the water w.r.t. the ground or commonly known as the ____________

River Crossing Problems River crossing usually involves a swimmer or boat crossing a river that has a “current” Bvw = boat’s velocity with respect to water the direction of Bvw is called the heading Heading is also the direction that the boat is pointed Bvg = boat’s velocity with respect to the ground or shore The direction of Bvg is the direction the boat actually moves with respect to a shore or ground observer Wvg = velocity of the water w.r.t. the ground or commonly known as the current

River Crossing Problems There is a vector chain rule equation we can create to relate Bvg , Bvw , and Wvg . What is it?

River Crossing Problems Bvg = Bvw + wvg

River Crossing Problems Bvg = Bvw + wvg Why is the heading (or the direction of Bvw or the direction the boat is pointed) not the same direction that the boat actually goes with respect to points on shore ( or the direction of Bvg ) ?

River Crossing Problems Bvg = Bvw + wvg Why is the heading (or the direction of Bvw or the direction the boat is pointed) not the same direction that the boat actually goes with respect to points on shore ( or the direction of Bvg ) ? The current can cause a boat or swimmer to move in a different direction w.r.t. shore.

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X has no current River Y has a current = 5.0 m/s [E] Boat X Start Boat Y start

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] After 1.0 seconds , where are the two boats? River X has no current River Y has a current = 5.0 m/s [E] Boat X Start Boat Y start

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] After 1.0 seconds , where are the two boats? River X has no current River Y has a current = 5.0 m/s [E] 5.0 m 20.0 m 20.0 m Boat X Start Boat Y start

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] After 2.0 seconds , where are the two boats? River X has no current River Y has a current = 5.0 m/s [E] 5.0 m 20.0 m 20.0 m Boat X Start Boat Y start

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] After 2.0 seconds , where are the two boats? River X has no current River Y has a current = 5.0 m/s [E] 5.0 m 20.0 m 20.0 m 5.0 m 20.0 m 20.0 m Boat X Start Boat Y start

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] After 3.0 seconds , where are the two boats? River X has no current River Y has a current = 5.0 m/s [E] 5.0 m 20.0 m 20.0 m 5.0 m 20.0 m 20.0 m Boat X Start Boat Y start

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] Note the two boats take the same time to cross the river. 5.0 m 20.0 m River X has no current River Y has a current = 5.0 m/s [E] 20.0 m 5.0 m 20.0 m 20.0 m 5.0 m 20.0 m 20.0 m Boat X Start Boat Y start

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] Note the two boats take the same time to cross the river. Boat X has a downstream displacement of _____ m [E] Downstream displacement ? 5.0 m 20.0 m River X has no current River Y has a current = 5.0 m/s [E] 20.0 m 5.0 m 20.0 m 20.0 m 5.0 m 20.0 m 20.0 m Boat X Start Boat Y start

Boat X and Boat Y both have Bvw = 20.0 m/s [N] River X and river Y both have the same width of 60.0 meters. Boat X and Boat Y start on the south shore at the same time. Which boat gets across the river to the north shore first? Boat X and Boat Y both have Bvw = 20.0 m/s [N] Note the two boats take the same time to cross the river. Boat X has a downstream displacement of 15.0 m [E] Downstream displacement ? 5.0 m 20.0 m River X has no current River Y has a current = 5.0 m/s [E] 20.0 m 5.0 m 20.0 m 20.0 m 5.0 m 20.0 m 20.0 m Boat X Start Boat Y start

Some river crossing rules!!

Some river crossing rules!! The time to cross a river only depends on the velocity component going _______________ the river.

Some river crossing rules!! The time to cross a river only depends on the velocity component going across the river.

Some river crossing rules!! The time to cross a river only depends on the velocity component going across the river. We will call the velocity Vacross

Some river crossing rules!! The time to cross a river only depends on the velocity component going across the river. We will call the velocity Vacross To calculate the time to cross, use the formula: t = w/Vacross where w is the river width and Vacross = lVacrossl

Some river crossing rules!! The downstream (or upstream) displacement depends only on the velocity component _______________ to the current.

Some river crossing rules!! The downstream (or upstream) displacement depends only on the velocity component parallel to the current.

Some river crossing rules!! The downstream (or upstream) displacement depends only on the velocity component parallel to the current. We will this velocity Vparallel

Some river crossing rules!! The downstream (or upstream) displacement depends only on the velocity component parallel to the current. We will this velocity Vparallel The formula for downstream or upstream displacement is: ∆dparallel = Vparallel t

Example #1: Tina can swim at 4. 00 m/s with respect to the water Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore

Example #1: Tina can swim at 4. 00 m/s with respect to the water Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore Given:

Example #1: Tina can swim at 4. 00 m/s with respect to the water Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore Given: TVW = ? TVg = ? WVg = ?

Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E]

Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula:

Tvg = Tvw + wvg Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg

Tvg = Tvw + wvg Sub: Given: TVW = 4.00 m/s [N] TVg = ? Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub:

Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Given: Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E]

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Given: Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] 4.00 m/s Tvg

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = ? Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = ? 4.00 m/s Tvg

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = (42 + 32)1/2 4.00 m/s Tvg

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = (42 + 32)1/2 = 5.00 m/s 4.00 m/s Tvg

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = (42 + 32)1/2 = 5.00 m/s Where is the reference angle? 4.00 m/s Tvg

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = (42 + 32)1/2 = 5.00 m/s Where is the reference angle? 4.00 m/s Tvg Θ

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = (42 + 32)1/2 = 5.00 m/s Θ = ? 4.00 m/s Tvg Θ

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = (42 + 32)1/2 = 5.00 m/s Θ = tan-1 (3/4) = 36.9° 4.00 m/s Tvg Θ

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = (42 + 32)1/2 = 5.00 m/s Θ = tan-1 (3/4) = 36.9° Tvg = ? 4.00 m/s Tvg Θ

Tvg Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore 3.00 m/s Given: TVW = 4.00 m/s [N] TVg = ? WVg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: Tvg =4.00 m/s [N] + 3.00 m/s [E] lTvgl = (42 + 32)1/2 = 5.00 m/s Θ = tan-1 (3/4) = 36.9° Tvg = 5.00 m/s [N 36.9°E] 4.00 m/s Tvg Θ

What formula gives us the time to cross? Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore What formula gives us the time to cross?

Example #1: Tina can swim at 4. 00 m/s with respect to the water Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore t = w/Vacross

Example #1: Tina can swim at 4. 00 m/s with respect to the water Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore t = w/Vacross = 80.0 m/4.00 m/s = ?

Example #1: Tina can swim at 4. 00 m/s with respect to the water Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore t = w/Vacross = 80.0 m/4.00 m/s = 20.0 s

Example #1: Tina can swim at 4. 00 m/s with respect to the water Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore t = w/Vacross = 80.0 m/4.00 m/s = 20.0 s What formula gives us the downstream/upstream displacement?

∆dparallel = Vparallel t = ? Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore t = w/Vacross = 80.0 m/4.00 m/s = 20.0 s ∆dparallel = Vparallel t = ?

∆dparallel = Vparallel t = 3.00 m/s [E] X 20.0 s Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore t = w/Vacross = 80.0 m/4.00 m/s = 20.0 s ∆dparallel = Vparallel t = 3.00 m/s [E] X 20.0 s

∆dparallel = Vparallel t = 3.00 m/s [E] X 20.0 s = 60.0 m [E] Example #1: Tina can swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) She is on the south side of a 80.0 m wide river that flows at 3.00 m/s [E]. She starts to cross the river to the north side by pointing herself north. Use GUFSA to find: a) Tina’s velocity with respect to a ground observer on the shore b) time to cross the river c) Tina’s downstream displacement when she arrives on the north shore t = w/Vacross = 80.0 m/4.00 m/s = 20.0 s ∆dparallel = Vparallel t = 3.00 m/s [E] X 20.0 s = 60.0 m [E]

Example #2: Tommy, Tina’s little brother, can also swim at 4 Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river

Example #2: Tommy, Tina’s little brother, can also swim at 4 Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given:

Example #2: Tommy, Tina’s little brother, can also swim at 4 Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = ? Tvw = ? Wvg = ?

Given: Tvg = _________ [N] Tvw = ? Wvg = ? Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = ? Wvg = ?

Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = ? Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = ?

Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E]

Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula:

Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg

Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub:

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E]

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] How can we draw a tip-to-tail diagram for this?

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] How can we draw a tip-to-tail diagram for this? Start by drawing the answer vector Tvg

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] How can we draw a tip-to-tail diagram for this? Start by drawing the answer vector Tvg Tvg

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] How can we draw a tip-to-tail diagram for this? The two vectors on the right side must give the answer vector Tvg . How should they be arranged? Tvg

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] How can we draw a tip-to-tail diagram for this? The two vectors on the right side must give the answer vector Tvg . Arrange them tip-to-tail but start with the known vector! Tvg

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river Given: Tvg = _________ [N] Tvw = 4.00 m/s [ ? ] Wvg = 3.00 m/s [E] Formula: Tvg = Tvw + wvg Sub: _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] How can we draw a tip-to-tail diagram for this? The two vectors on the right side must give the answer vector Tvg . Arrange them tip-to-tail but start with the known vector! Tvg 4.00 m/s 3.00 m/s

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Tvg 4.00 m/s 3.00 m/s

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] What angle defines Tommy’s heading ? Tvg 4.00 m/s 3.00 m/s

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] What angle defines Tommy’s heading ? Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = ? Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = ? Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = ? Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = ? Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° What is Tommy’s heading? Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° What is Tommy’s heading? Tommy must point himself [W41.4°N] Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° What is Tommy’s heading? Tommy must point himself [W41.4°N] The time to cross depends on the Vacoss How can we calculate Vacoss ? Vacoss = ? Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° What is Tommy’s heading? Tommy must point himself [W41.4°N] The time to cross depends on the Vacoss How can we calculate Vacoss ? Vacoss = (42 – 32)1/2 Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° What is Tommy’s heading? Tommy must point himself [W41.4°N] The time to cross depends on the Vacoss How can we calculate Vacoss ? Vacoss = (42 – 32)1/2 = 2.65 m/s Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° What is Tommy’s heading? Tommy must point himself [W41.4°N] The time to cross depends on the Vacoss How can we calculate Vacoss ? Vacoss = (42 – 32)1/2 time to cross formula = t = w/Vacross = 2.65 m/s Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° What is Tommy’s heading? Tommy must point himself [W41.4°N] The time to cross depends on the Vacoss How can we calculate Vacoss ? Vacoss = (42 – 32)1/2 time to cross formula = t = w/Vacross = 2.65 m/s t = 80.0 m/ 2.65m/s = ? Tvg 4.00 m/s 3.00 m/s Θ

_________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° Example #2: Tommy, Tina’s little brother, can also swim at 4.00 m/s with respect to the water. (sometimes stated as w.r.t. still water) He is on the south side of the same 80.0 m wide river that flows at 3.00 m/s [E]. He starts to cross the river to the north side by pointing himself so he will end up at a point on the north shore directly north of his starting point on the south shore. In other words, he uses a heading so there is no downstream displacement when he gets to the other side. Use GUFSA to find: a) the direction Tommy must point himself as he swims (i.e. his heading) b) time to cross the river _________ [N] = 4.00 m/s [ ? ] + 3.00 m/s [E] Θ = cos-1(3/4) = 41.4° What is Tommy’s heading? Tommy must point himself [W41.4°N] The time to cross depends on the Vacoss How can we calculate Vacoss ? Vacoss = (42 – 32)1/2 time to cross formula = t = w/Vacross = 2.65 m/s t = 80.0 m/ 2.65m/s = 30.2 seconds Tvg 4.00 m/s 3.00 m/s Θ

Try this hard one!! A boat is on the south shore of a 480.0 m wide river. The river’s current is 8.00 m/s [E]. The boat is pointed with a heading of [N36.9°W] and has a speed of 15.0 m/s with respect to the water. Use GUFSA to find: a) the velocity of the boat with respect to the ground or shore b) time to cross c) the upstream or downstream displacement when it reaches the north shore We will take it up in class!