Question

A deer walking at 3.29 m/s speeds up to 13.38 m/s, covering a distance of 63.5 meters during this acceleration. How much time did it take him to do this?

Answer 1

Assuming the deer's acceleration was constant, we have

$\Delta x=\dfrac{v_f+v_0}2t$

where $\Delta x$ is the deer's total displacement, $v_f,v_0$ are the deer's final and initial velocities, and $t$ is time. Then

$63.5\,\mathrm m=\dfrac{13.38\,\frac{\mathrm m}{\mathrm s}+3.29\,\frac{\mathrm m}{\mathrm s}}2t$

$\implies t=7.62\,\mathrm s$