A body starts from rest and uniformly accelerated to $12 \mathrm{~m} / \mathrm{s}$ in 5 seconds. Dis

Question

A body starts from rest and uniformly accelerated to $12 \mathrm{~m} / \mathrm{s}$ in 5 seconds. Distance covered during this process is

Professor PC Thomas Classes · Answer

Using $v = u+at \implies 12 = 0+a(5) \implies a = 2.4 \mathrm{m/s}^2$. Distance $s = ut+\frac{1}{2}at^2 = 0 + \frac{1}{2}(2.4)(5^2) = 30 \mathrm{m}$.