Multiplying a Vector by a Positive Scalar

Multiplying a Vector by a Positive Scalar

We represent a vector as an arrow. What happens to that arrow when we multiply it by a positive scalar?

We represent a vector as an arrow. What happens to that arrow when we multiply it by a positive scalar? Example: Consider vector V V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a positive scalar? Example: Consider vector V What does 3V look like? V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a positive scalar? Example: Consider vector V What does 3V look like? V 3V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a positive scalar? Example: Consider vector V What does 3V look like? What does (1/2)V look like? V 3V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a positive scalar? Example: Consider vector V What does 3V look like? What does (1/2)V look like? V (1/2)V 3V

Rule: Multiplying a Vector by a Positive Scalar
When multiplying a vector like velocity or acceleration by a positive scalar quantity like mass or time, the length or magnitude of the new vector ___________ but the ________________ remains the same.

When multiplying a vector like velocity or acceleration by a positive scalar quantity like mass or time, the length or magnitude of the new vector changes but the direction remains the same.

When multiplying a vector like velocity or acceleration by a positive scalar quantity like mass or time, the length or magnitude of the new vector changes but the direction remains the same. Vector equation Examples: Fnet= ma or a = ΔV/Δt

When multiplying a vector like velocity or acceleration by a positive scalar quantity like mass or time, the length or magnitude of the new vector changes but the direction remains the same. Vector equation Examples: Fnet= ma or a = ΔV/Δt Because of the rule, what can be said about Fnet and a ?

When multiplying a vector like velocity or acceleration by a positive scalar quantity like mass or time, the length or magnitude of the new vector changes but the direction remains the same. Vector equation Examples: Fnet= ma or a = ΔV/Δt Fnet and a are always in the same direction!

When multiplying a vector like velocity or acceleration by a positive scalar quantity like mass or time, the length or magnitude of the new vector changes but the direction remains the same. Vector equation Examples: Fnet= ma or a = ΔV/Δt Fnet and a are always in the same direction! a and ΔV are always in the same direction!

Multiplying a Vector by a Scalar = -1
We represent a vector as an arrow. What happens to that arrow when we multiply it by a scalar = -1? Example: Consider vector V V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a scalar = -1? Example: Consider vector V What does -1V or -V look like? V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a scalar = -1? Example: Consider vector V What does -1V or -V look like? V -1V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a scalar = -1? Example: Consider vector V What does -1V or -V look like? Rule: When we multiply a vector by a scalar =-1 or take the negative of a vector, the _____________ of the vector stays the same, but the _____________ becomes 180° opposite to the original vector. V -1V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a scalar = -1? Example: Consider vector V What does -1V or -V look like? Rule: When we multiply a vector by a scalar =-1 or take the negative of a vector, the magnitude of the vector stays the same, but the direction becomes 180° opposite to the original vector. V -1V

Multiplying a Vector by a Negative Scalar ≠ -1
We represent a vector as an arrow. What happens to that arrow when we multiply it by a negative scalar ≠ -1? Example: Consider vector V V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a negative scalar ≠ -1? Example: Consider vector V What does -2V look like? V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a negative scalar ≠ -1? Example: Consider vector V What does -2V look like? V -2V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a negative scalar ≠ -1? Example: Consider vector V What does -2V look like? What does (-1/2)V look like? V -2V

We represent a vector as an arrow. What happens to that arrow when we multiply it by a negative scalar ≠ -1? Example: Consider vector V What does -2V look like? What does (-1/2)V look like? V -2V (-1/2)V

Rule: Multiplying a Vector by a Negative Scalar ≠ -1
When multiplying a vector like velocity or acceleration by a negative scalar like charge, the length or magnitude of the new vector ___________ and the ________________ becomes 180° opposite to the original vector.

Rule: Multiplying a Vector by a Negative Scalar ≠ -1
When multiplying a vector like velocity or acceleration by a negative scalar like charge, the length or magnitude of the new vector changes and the direction becomes 180° opposite to the original vector.

Subtracting Vectors Consider vectors A and B A B

Subtracting Vectors Consider vectors A and B How do we arrange the
vectors to find the vector sum A + B ? A B

vectors to find the vector sum A + B ? B A + B A

vectors to find the vector difference A - B ? B A + B A

vectors to find the vector difference A - B ? We add the opposite vector A – B = A + (-B) B A + B A

vectors to find the vector difference A - B ? We add the opposite vector A – B = A + (-B) What does this mean? B A + B A

vectors to find the vector difference A - B ? We add the opposite vector A – B = A + (-B) We arrange A and –B tip-to tail! B A + B A

vectors to find the vector difference A - B ? We add the opposite vector A – B = A + (-B) We arrange A and –B tip-to tail! B A + B A -B A - B

Subtracting Vectors B A + B A -B A - B

Example: A car moving [E] maintains the same speed of 36
Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2.

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2 Given:

Given: V1 = 36.0 km/h [E] V2 = 36.0 km/h [N] ∆t = 2.00 seconds
Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] V2 = 36.0 km/h [N] ∆t = 2.00 seconds

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds Unknowns:

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds Unknowns: ∆V = ? a = ?

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds Unknowns: Formula: ∆V = ? a = ?

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds Unknowns: Formula: ∆v = ? ∆V = ? a = ?

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds Unknowns: Formula: ∆v = v2 – v1 ∆V = ? a = ?

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds Unknowns: Formula: ∆v = v2 – v1 ∆V = ? Substitute: a = ?

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds Unknowns: Formula: ∆v = v2 – v1 ∆V = ? Substitute: ∆v = 10.0 m/s [N] – 10.0 m/s [E] a = ?

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Given: V1 = 36.0 km/h [E] = 10.0 m/s [E] (divide by 3.6) V2 = 36.0 km/h [N] = 10.0 m/s [N] ∆t = 2.00 seconds Unknowns: Formula: ∆v = v2 – v1 ∆V = ? Substitute: ∆v = 10.0 m/s [N] – 10.0 m/s [E] a = ? ∆v = 10.0 m/s [N] m/s [W] This additional line must be shown on tests!!

∆v = 10.0 m/s [N] + 10.0 m/s [W] How do we find ∆v now?
Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] How do we find ∆v now?

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] Draw a tip-to-tail diagram of perpendicular vectors!

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] 10.0 m/s 10.0 m/s

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] 10.0 m/s 10.0 m/s ∆v

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] What is the reference angle that defines the direction of the vector? 10.0 m/s 10.0 m/s ∆v

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] 10.0 m/s 10.0 m/s ∆v θ

θ ∆v = 10.0 m/s [N] + 10.0 m/s [W] │∆v│= ?
Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ? 10.0 m/s 10.0 m/s ∆v θ

θ ∆v = 10.0 m/s [N] + 10.0 m/s [W] │∆v│= (102 + 102)1/2
Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ( )1/2 10.0 m/s 10.0 m/s ∆v θ

θ ∆v = 10.0 m/s [N] + 10.0 m/s [W] │∆v│= (102 + 102)1/2 │∆v│= 14.1 m/s
Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ( )1/2 │∆v│= 14.1 m/s 10.0 m/s 10.0 m/s ∆v θ

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ( )1/2 │∆v│= 14.1 m/s Ө = ? 10.0 m/s 10.0 m/s ∆v θ

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ( )1/2 │∆v│= 14.1 m/s Ө = tan-1 ( o/a) 10.0 m/s 10.0 m/s ∆v θ

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ( )1/2 │∆v│= 14.1 m/s Ө = tan-1 ( o/a) Ө = tan-1 ( 10/10) 10.0 m/s 10.0 m/s ∆v θ

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ( )1/2 │∆v│= 14.1 m/s Ө = tan-1 ( o/a) Ө = tan-1 ( 10/10) = 45° 10.0 m/s 10.0 m/s ∆v θ

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ( )1/2 │∆v│= 14.1 m/s Ө = tan-1 ( o/a) Ө = tan-1 ( 10/10) = 45° Answer: v = ? 10.0 m/s 10.0 m/s ∆v θ

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. ∆v = 10.0 m/s [N] m/s [W] │∆v│= ( )1/2 │∆v│= 14.1 m/s Ө = tan-1 ( o/a) Ө = tan-1 ( 10/10) = 45° Answer: Δv = 14.1 m/s [N45°W] 10.0 m/s 10.0 m/s ∆v θ

Answer: Δv = 14.1 m/s [N45°W] or [NW]
Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Answer: Δv = 14.1 m/s [N45°W] or [NW]

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Answer: Δv = 14.1 m/s [N45°W] or [NW] In terms of Δv , what formula will give us acceleration?

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Answer: Δv = 14.1 m/s [N45°W] or [NW] In terms of Δv , what formula will give us acceleration? a = Δv/ Δt

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Answer: Δv = 14.1 m/s [N45°W] or [NW] In terms of Δv , what formula will give us acceleration? a = Δv/ Δt = 14.1 m/s [NW] / 2.00 s

Example: A car moving [E] maintains the same speed of 36.0 km/h as it turns a corner in 2.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s b) the average acceleration in m/s2. Answer: Δv = 14.1 m/s [N45°W] or [NW] In terms of Δv , what formula will give us acceleration? a = Δv/ Δt = 14.1 m/s [NW] / 2.00 s = 7.05 m/s2 [NW]

Example #2: A car moving [E36. 9°S] maintains the same speed of 72
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2.

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given:

Given: V1 = 72.0 km/h [E36.9°S] V2 = 72.0 km/h [N] ∆t = 5.00 seconds
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: V1 = 72.0 km/h [E36.9°S] V2 = 72.0 km/h [N] ∆t = 5.00 seconds

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: V1 = 72.0 km/h [E36.9°S] =? V2 = 72.0 km/h [N] =? ∆t = 5.00 seconds

Given: V1 = 20.0 m/s [E36.9°S] V2 = 20.0 m/s [N] ∆t = 5.00 seconds
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: V1 = 20.0 m/s [E36.9°S] V2 = 20.0 m/s [N] ∆t = 5.00 seconds

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: V1 = 20.0 m/s [E36.9°S] V2 = 20.0 m/s [N] ∆t = 5.00 seconds Unknowns:

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: V1 = 20.0 m/s [E36.9°S] V2 = 20.0 m/s [N] ∆t = 5.00 seconds Unknowns: ∆V = ? a = ?

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: Formula: V1 = 20.0 m/s [E36.9°S] V2 = 20.0 m/s [N] ∆t = 5.00 seconds Unknowns: ∆V = ? a = ?

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: Formula: V1 = 20.0 m/s [E36.9°S] ∆v = ? V2 = 20.0 m/s [N] ∆t = 5.00 seconds Unknowns: ∆V = ? a = ?

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: Formula: V1 = 20.0 m/s [E36.9°S] ∆v = v2 – v1 V2 = 20.0 m/s [N] ∆t = 5.00 seconds Unknowns: ∆V = ? a = ?

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: Formula: V1 = 20.0 m/s [E36.9°S] ∆v = v2 – v1 V2 = 20.0 m/s [N] Substitute: ∆t = 5.00 seconds Unknowns: ∆V = ? a = ?

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: Formula: V1 = 20.0 m/s [E36.9°S] ∆v = v2 – v1 V2 = 20.0 m/s [N] Substitute: ∆t = 5.00 s ∆v = 20.0 m/s [N] – 20.0 m/s [E36.9°S] Unknowns: ∆V = ? a = ?

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: Formula: V1 = 20.0 m/s [E36.9°S] ∆v = v2 – v1 V2 = 20.0 m/s [N] Substitute: ∆t = 5.00 s ∆v = 20.0 m/s [N] – 20.0 m/s [E36.9°S] Unknowns: = ? ∆V = ? a = ?

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. Given: Formula: V1 = 20.0 m/s [E36.9°S] ∆v = v2 – v1 V2 = 20.0 m/s [N] Substitute: ∆t = 5.00 s ∆v = 20.0 m/s [N] – 20.0 m/s [E36.9°S] Unknowns: ∆v = 20.0 m/s [N] m/s [W36.9°N] ∆V = ? a = ?

∆v = 20.0 m/s [N] + 20.0 m/s [W36.9°N] How do we find the vector sum?
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] How do we find the vector sum?

∆v = 20.0 m/s [N] + 20.0 m/s [W36.9°N] Tip-to tail diagram ?
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ?

∆v = 20.0 m/s [N] + 20.0 m/s [W36.9°N] Tip-to tail diagram ?
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? 20m/s 36.9° ∆v 20 m/s

∆v = 20.0 m/s [N] + 20.0 m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀ = ?
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀ = ? 20m/s 36.9° ∆v 20 m/s

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) 20m/s 36.9° ∆v 20 m/s

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s 20m/s 36.9° ∆v 20 m/s

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s What is the reference angle? 20m/s 36.9° ∆v 20 m/s

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s What is the reference angle? 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s Ө = ? 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s Ө = Use sin law or isosceles geometry 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s Ө = Use sin law or isosceles geometry sinӨ/20 = sin (126.9)/35.8 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s Ө = Use sin law or isosceles geometry sinӨ/20 = sin (126.9)/35.8 Ө = 26.5° or … 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s Ө = Use sin law or isosceles geometry sinӨ/20 = sin (126.9)/35.8 Ө = 26.5° or …(180° )/2 = 26.6° 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s Ө = Use sin law or isosceles geometry sinӨ/20 = sin (126.9)/35.8 Ө = 26.5° or …(180° )/2 = 26.6° ∆v = ??? 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 20.0 m/s [N] m/s [W36.9°N] Tip-to tail diagram ? ׀∆v ׀2 = –2(20)(20)cos(126.9°) ׀∆v׀ = 35.8 m/s Ө = Use sin law or isosceles geometry sinӨ/20 = sin (126.9)/35.8 Ө = 26.5° or …(180° )/2 = 26.6° ∆v = 35.8 m/s [N 26.6°W] 20m/s 36.9° ∆v 20 m/s Ө

∆v = 35.8 m/s [N 26.6°W] ∆t = 5.00 seconds a = formula ???
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 35.8 m/s [N 26.6°W] ∆t = 5.00 seconds a = formula ??? 20m/s 36.9° ∆v 20 m/s Ө

∆v = 35.8 m/s [N 26.6°W] ∆t = 5.00 seconds a = ∆v / ∆t
Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 35.8 m/s [N 26.6°W] ∆t = 5.00 seconds a = ∆v / ∆t 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 35.8 m/s [N 26.6°W] ∆t = 5.00 seconds a = ∆v / ∆t = 35.8 m/s [N 26.6°W]/ 5.00s 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 35.8 m/s [N 26.6°W] ∆t = 5.00 seconds a = ∆v / ∆t = 35.8 m/s [N 26.6°W]/ 5.00s = 7.16 m/s2 [N 26.6°W] 20m/s 36.9° ∆v 20 m/s Ө

Example #2: A car moving [E36.9°S] maintains the same speed of 72.0 km/h as it turns a corner in 5.00 seconds. At the end of the turn, the car is moving north. Using GUFSA, find: a) the change-in velocity in m/s (use cos/sin law) b) the average acceleration in m/s2. ∆v = 35.8 m/s [N 26.6°W] ∆t = 5.00 seconds a = ∆v / ∆t = 35.8 m/s [N 26.6°W]/ 5.00s = 7.16 m/s2 [N 26.6°W] Now try #4 on lesson 8 exercises! We will put a solution on the board. 20m/s 36.9° ∆v 20 m/s Ө

Multiplying a Vector by a Positive Scalar

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