A car travelling at 22.4m/s skids to stop in 2.50s. Determine the skidding distance of the car and its uniform acceleration.

Hello!

The acceleration (actually, deceleration) of a car is assumed to be constant. This is more or less correct because of Newton's Second law, `a = F/m,` and the breaking force `F` is more or less constant.

Denote the magnitude of acceleration as `a` (a positive number). Also denote the initial speed as `V_0` and the time of a stop as `t_1.`

The speed `V` changes uniformly with a time, the formula is obviously `V(t) = V_0 - a t.`  The time `t_1` when a car is stopped is that moment when `V(t_1) = 0,` i.e. `V_0 = a t_1` and `a = V_0/t_1 = 22.4 / 2.50 = 8.96 (m/s^2).`

The distance `D` after the skidding begins is `D(t) = V_0 t - (a t^2)/2,` so the total skidding distance is `D(t_1) = V_0 t_1 - (V_0/t_1 *t_1^2)/2 = (V_0 t_1)/2 = 28 (m).`

So the answers: the skidding distance is 28 m and the uniform acceleration is 8.96 `m/s^2.`