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?

Question

asked Jan 21, 2019 186k views

1 Answer

Pankaj Saha · Answer 1 · 2019-01-25T12:55:40+0000

Assuming the deer's acceleration was constant, we have

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

where
$\Delta x$ is the deer's total displacement,
v_f,v_0 are the deer's final and initial velocities, and
is time. Then

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

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