Physics - Kinematics Question with Solution | TestHub

The distance traveled in the $n$-th second is given by $S_n=u+\frac{a}{2}(2n-1)$.

Given $u=0$ (starts from rest) and $S_{10}=19$ m for the tenth second ($n=10$).

So, $19=0+\frac{a}{2}(2\times 10-1)$.

$19=\frac{a}{2}(19)$.

$a=2$ m/s².

The final answer is $\boxed{2m/s^2}$.