Question

A car moving at 30m/s slows down uniformly to a speed of 10m/s in a time of 5.0s. Determine a) the acceleration of the car, and b) the distance out moved in the third second

Answer 1

Answer:

DATA:

v1 = 30m/s

v2 = 10 m/s

a = ?

t = 5 s

d = ?

SOLUTION:

Solving for the acceleration of the car

a = (10 m/s-30 m/s) / 5s

a = -20 m/s / 5s

a = -4 m/s^2

Solving for the distance traveled after the third second

d = v1 * t + 1/2at^2

d = 30 m/s * 3 s + -2m/s^2 * (3s)^2

d = 90 m + - 18 m

d = 72 m

Solving for the distance traveled after 2 seconds

d = v1 * t + 1/2 at^2

d = 10 m/s * 2 s + -2 m/s^2 * (2 s)^2

d = 20 m + - 8 m

d = 12 m

Solving for the distance traveled in the third second

d = distance moved after 3 s – distance moved after 2 s

d = 72 m – 12 m

d = 60 m

Answer 2

Answer:

accleration = -4m/s

distance = 20m