A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.89 m/s^2 for t1 = 20 s. At that point the driver notices a tree limb that has fallen on the road and brakes hard for t2 s until it comes to a stop. Required:a. Write an expression for the car's speed just before the driver begins braking, v1.b. If the limb is on the road at a distance of 550 meters from where the car began its initial acceleration, will the car hit the limb?c. How far, in meters, from the original location of the limb will the car be when it stops?

A.) 37.8 m/s

B.) No

C.) 172 m

Explanation:

Given that  car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.89 m/s^2 for t1 = 20 s. At that point the driver notices a tree limb that has fallen on the road and brakes hard for t2 s until it comes to a stop. Required:

a.) Write an expression for the car's speed just before the driver begins braking, v1.

Since the car accelerate from rest, initial velocity U = 0

Using first equation of motion

V1 = U + at

V1 = 1.89 × 20

V1 = 37.8 m/s

b.) If the limb is on the road at a distance of 550 meters from where the car began its initial acceleration, will the car hit the limb?

Using third equation of motion

V^2 = U^2 + 2as

37.8^2 = 0 + 2 × 1.89 × S

1428.84 = 3.78S

S = 1428.84 / 3.78

S = 378m

Since the limb is on the road at a distance of 550 meters from where the car began its initial acceleration, we can therefore conclude that the car will not hit the limb.

c. How far, in meters, from the original location of the limb will the car be when it stops?

550 - 378 = 172 m

The car will be 172 metres from the limb.

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