Question

A car pulls away from an intersection when the light turns green. After uniformly accelerating for the next 6.0 s, the car has traveled a distance of 75 m. The car then proceeds at a constant speed. ( /7)

(a) What was the car's acceleration?
(b) How fast was the car traveling when it finished accelerating?
(c) How long will it take for the car to travel another 85 m at this constant speed?

Answer 1

Hi there!

(a)

We can use the kinematic equation:

d = vit + 1/2at²

vi = initial velocity (0 m/s since the car starts from rest)

a = acceleration (m/s²)

t = time (sec)

Rearrange the equation to solve for acceleration:

d = 1/2at²

2d/t² = a

2(75)/(6²) = 150/36 = 4.167 m/s²

(B)

Now, that we know the car's acceleration, we can solve for its final velocity:

vf = vi + at (initial velocity = 0 m/s)

vf = at

vf = (4.167)(6) = 25 m/s

(C)

Now, we can use the equation:

distance = speed × time, or d = vt

Rearrange to solve for time:

d/v = t

85/25 = 3.4 sec