Answer
(a) $17.8m/s$
(b) $168m$
Work Step by Step
(a) We can find the required speed as follows:
$a=\frac{F}{m}$
We plug in the known values to obtain:
$a=\frac{-4.30\times 10^5}{3.50\times 10^5}=-1.23m/s$
Now $v=v_{\circ}+at$
We plug in the known values to obtain:
$v=27+(-1.23)(7.50)=17.8m/s$
(b) We can find the required distance as follows:
$\Delta x=\frac{1}{2}(v_{\circ}+v)\Delta t$
We plug in the known values to obtain:
$\Delta x=\frac{1}{2}(27.0+17.8)(7.50)$
$\Delta x=168m$
