· Firstly, we need to note down what we have: distance travelled, s =15 m; initial velocity, u = 28 ms^{-1}; final velocity, v = 0 because the car comes to rest; we are looking for the deceleration hence a; since we are using s, u, v and a, we need to use the "suvat" equation that contains those four letters ----- v^{2} = u^{2} + 2as· By rearranging the equation ---- a = (v^{2} - u^{2}) / 2s = -28^{2} / (2 x 15) = -784/30 = -26 ms^{-2} (the minus sign is because deceleration is a decrease in speed rather than an increase)