Question

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 with a constant acceleration of a2 = -5.93 m/s2.Write an expression for the car's speed just before the driver begins braking, v1.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?

Answer 1

Answer:

1) an expression for the car's speed is given as

v = u + at

where

v is the car's speed

u is the initial speed of the car

a is the car's acceleration

t is the time spent accelerating

2) The car does not hit the tree limb

Explanation:

The initial velocity of the car = 0 m/s  (since it accelerates from rest)

acceleration of the car = 1.76 m/s

time spent accelerating = 20 s

For the car's speed just before the driver begins braking, we use the expression

v = u + at

where v is the final speed of the car just before the driver begins braking

u is the initial velocity with which the car starts moving

a is the acceleration of the car

t is the time spent accelerating from u to v

substituting values, we have

v = 0 + 1.76(20)

v = 0 + 35.2

the car's speed v = 35.2 m/s

In this time the car accelerates, the car moves a distance given by

s = ut + $\frac{1}{2}$a$t^2$

where s is the distance covered in this time

u is the initial speed of the vehicle

a is the acceleration

t is the time taken

substituting, we have

s = 0(20) + $\frac{1}{2}$(1.76)$20^{2}$

s = 0 + 352

distance s = 352 m

When the driver brakes, we have

time spent braking = 5 s

acceleration = -5.93 m/s

and the distance to the limb = 550 m from where the car begun

to get the distance covered in this period, we use the expression

s = ut + $\frac{1}{2}$a$t^2$

where s is the distance traveled at this time

u is the speed of the car before it starts braking = 35.2 m/s

a is the acceleration at this point

t is the time taken to decelerate to a stop

substituting values, we have

s = 35.2(5) + $\frac{1}{2}$(-5.93 x $5^2$)

s = 176 - 74.125

s = 101.88 m

Total distance moved by the car = 352 m + 101.88 m = 453.88 m

Since the total distance traveled by the car is less than the distance from the starting point to the place where the tree limb is, the car does not hit the tree limb.