Question

A monkey driving a bananamobile spots a banana split 200 m ahead on the highway. The monkey uniformily decelerates for 5.0 s, stopping just in time to avoid squashing the banana split. Find (A) the initial velocity and (B) the acceleration of the bananamobile.

Answer 1

Equation of motion is

$200=vt-\frac{at^{2}}{2}$

where v is initial speed of bananamobile, $t=5$ is time of decelerating, and $a$ is acceleration of the bananamobile.

Equation of speed is

$v=at$

From this two equations we find that

$200=\frac{at^{2}}{2}$

$a=\sqrt{\frac{200\cdot 2}{t^{2}}}=\sqrt{\frac{400}{25}}=\sqrt{16}=4\,m/s^{2}$

The initial speed is

$at=4\cdot 5=20m/s$