Question

Scooby Doo is initially at rest when he sees a box of Scooby Snacks 55 m up ahead. If Scooby Doo accelerates at 0.44 m/s2, how long does it take him to reach the box of Scooby Snacks? 

Answer 1

The distance travelled under a constant acceleration from rest is:

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

where $d = 55m$, $a = 0.44m/s^2$, and $t$ is the time of motion. Expressing the time, obtain:

$$t = \sqrt{\dfrac{2d}{a}}\\ t = \sqrt{\dfrac{2\cdot 55}{0.44}} \approx 15.81s$$

Answer. 15.81s.