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A car traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.00 s later. (a) What is the magnitude of the car’s constant acceleration before impact? (b) How fast is the car traveling at impact?

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Answer: (A) acceleration = -3.6m/s². This is deceleration or negative acceleration.

(B) Therefore, velocity just before impact, v = 8.4m/s

Step-by-step explanation:

(a) Initial velocity, u = 56km/h = (56*1000/3600)m/s = 15.6m/s

Distance, s = 24.0m, time, t = 2.0s, acceleration, a = ?

Using the equation, s = ut + 1/2at²

24.0m = 15.6m/s * 2s + (a * 2²)/2

24.0 = 31.2 +2a

2a = -31.2 + 24

2a = -7.2

a = -7.2/2

acceleration = -3.6m/s². This is deceleration or negative acceleration.

(b) final velocity just before impact, v = ? initial velocity, u = 15.6m/s, acceleration = -3.6m/s², time, t = 2.0s

using the equation v = u + at

v = 15.6m/s + (-3.6m/s² * 2s)

v = 15.6m/s - 7.2m/s

v = 8.4m/s

Therefore, velocity just before impact, v = 8.4m/s

User Graham
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