Examples of 1-d Problems using the BIG FIVE or the Little One

Examples of 1-d Problems using the BIG FIVE or the Little One

Problem-Solving Strategies

Problem-Solving Strategies Use G U F S A

Problem-Solving Strategies Use G Write the given info in symbol form sometimes with a picture or diagram U F S A

Problem-Solving Strategies Use G Write the given info in symbol form sometimes with a picture or diagram U Write the unknown or intermediate unknowns in symbol form F S A

Problem-Solving Strategies Use G Write the given info in symbol form sometimes with a picture or diagram U Write the unknown or intermediate unknowns in symbol form F Look at the given and unknown(s) and decide what formula or formulas to write down S A

Problem-Solving Strategies Use G Write the given info in symbol form sometimes with a picture or diagram U Write the unknown or intermediate unknowns in symbol form F Look at the given and unknown(s) and decide what formula or formulas to write down S Substitute the given into your formulas, then show steps to get a final value. Watch units! A

Problem-Solving Strategies Use G Write the given info in symbol form sometimes with a picture or diagram U Write the unknown or intermediate unknowns in symbol form F Look at the given and unknown(s) and decide what formula or formulas to write down S Substitute the given into your formulas, then show steps to get a final value. Watch units! A Write the answers with units and/or direction

Side Notes on Problem-Solving Strategies

Side Notes on Problem-Solving Strategies You may have to find an intermediate unknown first using GUFSA, then substitute the intermediate unknown in a another formula to solve for your final unknown

Side Notes on Problem-Solving Strategies You may have to find an intermediate unknown first using GUFSA, then substitute the intermediate unknown in a another formula to solve for your final unknown If you have two objects in a problem, you may have to apply GUFSA separately for each object to form two equations with two unknowns, then solve

Side Notes on Problem-Solving Strategies You may have to find an intermediate unknown first using GUFSA, then substitute the intermediate unknown in a another formula to solve for your final unknown If you have two objects in a problem, you may have to apply GUFSA separately for each object to form two equations with two unknowns, then solve If you are still stuck, re-read and look for a simplifying physics principle or a different perspective to solve. Is there a mathematical relationship between variables?

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector a) 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector a) + - 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector a) Δd = -80.0 a = -10.0 v1 = +30.0 + - 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector a) Δd = Unknown: a = -10.0 v1 = +30.0 + - 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector a) Δd = Unknown: a = Δt = ? v1 = +30.0 + - 80.0m Ground level

a) Δd = -80.0 a = -10.0 v1 = +30.0 Δt = ?

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt - aΔt2/2

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula:

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute:

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -80 = 30t -10t2/2

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -80 = 30t -10t2/2 -80 = 30t -5t2

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -80 = 30t -10t2/2 -80 = 30t -5t2 5t2 – 30t -80 = 0

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -80 = 30t -10t2/2 -80 = 30t -5t2 5t2 – 30t -80 = 0 t2 – 6t -16 = 0

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 Substitute: -80 = 30t -10t2/2 -80 = 30t -5t2 5t2 – 30t -80 = 0 t2 – 6t -16 = 0

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 t=8 or -2 Substitute: -80 = 30t -10t2/2 -80 = 30t -5t2 5t2 – 30t -80 = 0 t2 – 6t -16 = 0

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 t=8 or -2 Substitute: -80 = 30t -10t2/2 t = ( - b ± √(b2- 4ac) ) / ( 2a ) -80 = 30t -5t2 5t2 – 30t -80 = 0 t2 – 6t -16 = 0

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 t=8 or -2 Substitute: -80 = 30t -10t2/ t = ( - b ± √(b2- 4ac) ) /(2a) -80 = 30t -5t t = ( + 6 ± √((-6)2-4(1)(-16)) / (2(1)) 5t2 – 30t -80 = 0 t2 – 6t -16 = 0

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 t=8 or -2 Substitute: -80 = 30t -10t2/2 t = ( - b ± √(b2- 4ac) ) /(2a) -80 = 30t -5t t = ( + 6 ± √((-6)2-4(1)(-16)) / (2(1)) 5t2 – 30t -80 = t = ( + 6 ± √(36 +64)) / (2) t2 – 6t -16 = 0

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 t=8 or -2 Substitute: -80 = 30t -10t2/2 t = ( - b ± √(b2- 4ac) ) /(2a) -80 = 30t -5t t = ( + 6 ± √((-6)2-4(1)(-16)) / (2(1)) 5t2 – 30t -80 = t = ( + 6 ± √(36 +64)) / (2) t2 – 6t -16 = t = ( + 6 ± √(100)) / (2)

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 t=8 or -2 Substitute: -80 = 30t -10t2/2 t = ( - b ± √(b2- 4ac) ) /(2a) -80 = 30t -5t t = ( + 6 ± √((-6)2-4(1)(-16)) / (2(1)) 5t2 – 30t -80 = t = ( + 6 ± √(36 +64)) / (2) t2 – 6t -16 = t = ( + 6 ± √(100)) / (2) t=8 or -2

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 t=8 or -2 Substitute: -80 = 30t -10t2/2 t = ( - b ± √(b2- 4ac) ) /(2a) -80 = 30t -5t t = ( + 6 ± √((-6)2-4(1)(-16)) / (2(1)) 5t2 – 30t -80 = t = ( + 6 ± √(36 +64)) / (2) t2 – 6t -16 = t = ( + 6 ± √(100)) / (2) t=8 or -2 Answer:

a) Δd = BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v1 = #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δt = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/ (t – 8) (t + 2) = 0 t=8 or -2 Substitute: -80 = 30t -10t2/2 t = ( - b ± √(b2- 4ac) ) /(2a) -80 = 30t -5t t = ( + 6 ± √((-6)2-4(1)(-16)) / (2(1)) 5t2 – 30t -80 = t = ( + 6 ± √(36 +64)) / (2) t2 – 6t -16 = t = ( + 6 ± √(100)) / (2) t=8 or -2 Answer: As time cannot be negative, time of flight = 8.00 seconds

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector b) + - 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector b) v1= +30.0 a = v2= 0.00 + - 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector b) v1= +30.0 a = v2= 0.00 Unknown(s): + - 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector b) v1= +30.0 a = v2= 0.00 Unknown(s): Δd = ? and max height = ? + - 80.0m Ground level

Easy Problem Example: A lady on a clifftop tosses a penny directly up at 30.0 m/s. When the penny leaves her hand, it is 80.0 m above the ground below. Neglecting air friction, determine: a) time of flight to the ground below b) maximum height above ground Given: green=scalar black= vector b) v1= +30.0 a = v2= 0.00 Unknown(s): Δd = ? and max height = ? + Δd - 80.0m Ground level

b) v1= +30.0 a = v2= 0.00 Δd = ? and max height = ?

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula:

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a)

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) Substitute:

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) Substitute: Δd = (02 – 302)/(2(-10))

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) Substitute: Δd = (02 – 302)/(2(-10)) Δd = (-900)/(-20)

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) Substitute: Δd = (02 – 302)/(2(-10)) Δd = (-900)/(-20) Δd = 45.0

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) Substitute: Δd = (02 – 302)/(2(-10)) Δd = (-900)/(-20) Δd = 45.0 Δd = m [up]

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) max height = ? Substitute: Δd = (02 – 302)/(2(-10)) Δd = (-900)/(-20) Δd = 45.0 Δd = m [up]

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) max height = ? Substitute: Δd = (02 – 302)/(2(-10)) = Δd = (-900)/(-20) Δd = 45.0 Δd = m [up]

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) max height = ? Substitute: Δd = (02 – 302)/(2(-10)) = Δd = (-900)/(-20) = 125.0 Δd = 45.0 Δd = m [up]

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) max height = ? Substitute: Δd = (02 – 302)/(2(-10)) = Δd = (-900)/(-20) = 125.0 Δd = 45.0 Δd = m [up] Answer:

b) v1= BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 v2= #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) Δd = ? max height = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = (v22 – v12)/(2a) max height = ? Substitute: Δd = (02 – 302)/(2(-10)) = Δd = (-900)/(-20) = 125.0 Δd = Answer: Δd = m [up] The maximum height of the penny above ground is m

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below?

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? There are two types of motion in this example. What are they?

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? There are two types of motion in this example. What are they? Constant acceleration of the stone down the well Constant speed of sound up the well back to the man

Harder Problem: A man drops a stone from rest down a well
Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant acceleration of the stone down the well

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant acceleration of the stone down the well Given:

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant acceleration of the stone down the well Given: v1= 0.00 m/s a = m/s/s (- means down)

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant acceleration of the stone down the well Given: v1= 0.00 m/s a = m/s/s ( - means down) Unknowns:

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant acceleration of the stone down the well Given: v1= 0.00 m/s a = m/s/s ( - means down) Unknowns: Δd = ? t = ?

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant acceleration of the stone down the well Given: v1= 0.00 m/s a = m/s/s ( - means down) Unknowns: Δd = ? t = ? This is not equal to 4.23 s, but only the time for the stone to fall down the well.

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s a = m/s/s Δd = ? t = ?

v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a
Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula:

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute:

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -d = 0 t -10 t2/2

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -d = 0 t -10 t2/2 -d = -10 t2/2

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -d = 0 t -10 t2/2 -d = -10 t2/2 d = 5 t2

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -d = 0 t -10 t2/2 -d = -10 t2/2 d = 5 t2 Equation 1

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? v1= 0.00 m/s BIG FIVE: #1a a = (v2 – v1)/Δt #1b Δt = (v2 – v1)/a a = m/s/s #1c v2 = v1 + aΔt #2 Δd = (v2 + v1)Δt/2 Δd = ? #3 Δd = v1Δt + aΔt2/ #4a Δd = (v22 – v12)/(2a) t = ? #4b v22 – v12 = 2aΔd #5 Δd = v2Δt – aΔt2/2 Formula: Δd = v1Δt + aΔt2/2 Substitute: -d = 0 t -10 t2/2 -d = -10 t2/2 d = 5 t2 Equation 1 Again, t represents only the time for the stone to fall down the well, not the total time of 4.23 seconds which includes the time for the stone to fall and the time for the sound to come back to the man's ear !!.

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: - down

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: v = m/s - down

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: v = m/s - down Unknown(s):

time for the sound to travel from water surface to the man's ear = ?
Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: v = m/s - down Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ?

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: v = m/s - down Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ? Can we express this time in terms of the time for the stone to fall, “t”?

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: v = m/s - down Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ? Can we express this time in terms of the time for the stone to fall, “t”? t = (time for sound to travel) = ?

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear up Given: v = m/s - down Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ? Can we express this time in terms of the time for the stone to fall, “t”? t = (time for sound to travel) = t

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: Formula: v = m/s down Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ? Can we express this time in terms of the time for the stone to fall, “t”? t = (time for sound to travel) = t

Given: Formula: Little one Δd = v t v = + 342 m/s - down Unknown(s):
Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear up Given: Formula: Little one Δd = v t v = m/s - down Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ? Can we express this time in terms of the time for the stone to fall, “t”? t = (time for sound to travel) = t

Given: Formula: Little one Δd = v t v = + 342 m/s - down Substitute:
Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: Formula: Little one Δd = v t v = m/s - down Substitute: Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ? Can we express this time in terms of the time for the stone to fall, “t”? t = (time for sound to travel) = t

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: Formula: Little one Δd = v t v = m/s - down Substitute:+d = +342 ( 4.23 – t ) Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ? Can we express this time in terms of the time for the stone to fall, “t”? t = (time for sound to travel) = t

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? Constant velocity of the sound from the water surface back up to the man's ear +up Given: Formula: Little one Δd = v t v = m/s down Substitute: +d = +342 ( 4.23 – t ) Equation 2 Unknown(s): Δd = +d time for the sound to travel from water surface to the man's ear = ? Can we express this time in terms of the time for the stone to fall, “t”? t = (time for sound to travel) = t

d = 5 t2 Equation 1 +d = +342 ( 4.23 – t ) Equation 2
Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution:

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: 5 t2 = ( 4.23 – t )

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: 5 t2 = ( 4.23 – t ) 5 t2 = – 342 t

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: 5 t2 = ( 4.23 – t ) 5 t2 = – 342 t 5 t t = 0

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: 5 t2 = ( 4.23 – t ) 5 t2 = – 342 t 5 t t = 0 t = ( - b ± √(b2- 4ac) ) / ( 2a )

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: 5 t2 = ( 4.23 – t ) 5 t2 = – 342 t 5 t t = 0 t = ( - b ± √(b2- 4ac) ) / ( 2a ) = ( ± √( (5)(-1450) ) / ( 2(5))

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: 5 t2 = ( 4.23 – t ) 5 t2 = – 342 t 5 t t = 0 t = ( - b ± √(b2- 4ac) ) / ( 2a ) = ( ± √( (5)(-1450) ) / ( 2(5)) t = 4.0 seconds or – 7.2 seconds

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: Back sub in either equation: 5 t2 = ( 4.23 – t ) 5 t2 = – 342 t 5 t t = 0 t = ( - b ± √(b2- 4ac) ) / ( 2a ) = ( ± √( (5)(-1450) ) / ( 2(5)) t = 4.0 seconds or – 7.2 seconds

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: Back sub in either equation: 5 t2 = ( 4.23 – t ) d = 5 t2 5 t2 = – 342 t 5 t t = 0 t = ( - b ± √(b2- 4ac) ) / ( 2a ) = ( ± √( (5)(-1450) ) / ( 2(5)) t = 4.0 seconds or – 7.2 seconds

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: Back sub in either equation: 5 t2 = ( 4.23 – t ) d = 5 t2 5 t2 = – 342 t d = 5 (4.0)2 5 t t = 0 t = ( - b ± √(b2- 4ac) ) / ( 2a ) = ( ± √( (5)(-1450) ) / ( 2(5)) t = 4.0 seconds or – 7.2 seconds

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: Back sub in either equation: 5 t2 = ( 4.23 – t ) d = 5 t2 5 t2 = – 342 t d = 5 (4.0)2 5 t t = 0 d = 80 or 8.0 X 101 m t = ( - b ± √(b2- 4ac) ) / ( 2a ) = ( ± √( (5)(-1450) ) / ( 2(5)) t = 4.0 seconds or – 7.2 seconds

Harder Problem: A man drops a stone from rest down a well. The man hears a splash 4.23 seconds later. Given the speed of sound in air is constant in the well at 342 m/s, how far did the stone fall down the well to the water surface below? d = 5 t2 Equation d = +342 ( 4.23 – t ) Equation 2 Solve by substitution: Back sub in either equation: 5 t2 = ( 4.23 – t ) d = 5 t2 5 t2 = – 342 t d = 5 (4.0)2 5 t t = 0 d = 80 or 8.0 X 101 m Answer : The well is 8.0 X 101 m deep. t = ( - b ± √(b2- 4ac) ) / ( 2a ) = ( ± √( (5)(-1450) ) / ( 2(5)) t = 4.0 seconds or – 7.2 seconds

Examples of 1-d Problems using the BIG FIVE or the Little One

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Examples of 1-d Problems using the BIG FIVE or the Little One

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Presentation on theme: "Examples of 1-d Problems using the BIG FIVE or the Little One"— Presentation transcript:

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