Answer to Question #275339 in Physics for Ash

Question

A car pulls away from an intersection when the light turns green. After uniformly accelerating for the next 4.0 seconds, the car has travelled a distance of 25 m. The car then proceeds at constant speed.

a) what was the cars acceleration?

b) how fast was the car travelling when it finished accelerating?

c) how long will it take for the car to travel another 25 m at this constant speed?

Answer 1

(a)

$$d=\dfrac{1}{2}at^2,$$ $$a=\dfrac{2d}{t^2}=\dfrac{2\times25\ m}{(4\ s)^2}=3.125\ \dfrac{m}{s^2}.$$

(b)

$$v=v_0+at=0+3.125\ \dfrac{m}{s^2}\times4\ s=12.5\ \dfrac{m}{s}.$$

(c)

$$t=\dfrac{d}{v}=\dfrac{25\ m}{12.5\ \dfrac{m}{s}}=2\ s.$$