A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the driver notices a tree limb that has fallen on the road and brakes hard for t2 = 5 s until it comes to a stop.

Part (a) Write an expression for the car's speed just before the driver begins braking, v1.
Part (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?
Part (c) How far, in meters, from the original location of the limb will the car be when it stops?

(a)

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

(b) The velocity of the car before the driver begins braking is

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

We can use the following equation of motion to calculate how far the car has travel since braking to stop

Also the distance from start to where the driver starts braking is

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

lol

explanation:

there are no components of a physical goal mentioned on the list of choices that you've provided.