Question

5. A soapbox derby race car starts at rest and moves 300 m down a track in 22.4 s. Calculate its acceleration (assumed constant) and its speed at the end of the track.

Answer 1

The distance travelled in time $t$ under the constant acceleration (and with zero initial speed) is:

$$d = \dfrac{at^2}{2}$$

where $a$ is the acceleration and $t = 22.4s$ is the time. If the distance is $d = 300m$, then expressing $a$, obtain:

$$a = \dfrac{2d}{t^2} = \dfrac{2\cdot200m }{(22.4s)^2} \approx 0.8m/s^2$$

The speed at the end of the track is:

$$v = \dfrac{2d}{t^2}t = \dfrac{2d}{t}\\ v = \dfrac{2\cdot 200m}{22.4s} \approx 17.9m/s$$

Answer. 0.8 m/s^2, 17.9 m/s.