# Question b0b36

We use one of the kinematic equations, ${v}^{2} = {u}^{2} + 2 a . s$ and, because we know that the initial velocity, u is zero, it simplifies to ${v}^{2} = 2 a . s$ so a = v^2/(2s#
$a = {559}^{2} / \left[2 \times 0.803\right)$
$a = 194570.9 \frac{m}{s} ^ 2$ as you are given 3 s.f. in the question we leave it as 3 s.f. so $a = 1.94 \times {10}^{5} m {s}^{-} 2$