Answer
(a) $\frac{v^2}{2\mu g}$
(b) quadrupled
(c) same
Work Step by Step
(a) We can find the required distance as follows:
$\Sigma F_x=-\mu N$
$\implies ma=-\mu mg$
We know that
$d=\frac{v_f^2-v^2}{2a}$
$\implies d=\frac{(0)^2-v^2}{2(-\mu g)}$
$\implies d=\frac{v^2}{2\mu g}$
(b) As from part (a) $ d=\frac{v^2}{2\mu g}$, this equation shows that the stopping distance is directly proportional to the square of the initial speed. Thus, the stopping distance will be quadrupled, if the speed is doubled.
(c) As from part (a) $ d=\frac{v^2}{2\mu g}$, this equation shows that the stopping distance does not depend on the mass. Thus, even if the mass is doubled, the stopping distance will remain the same.
