Answer to Question #213281 in Physics for saint

Question

A car travels between two stations, A and B, 4 km from each other. It starts from rest at A and accelerates uniformly until it obtains a speed of 80 km/h. This car maintains until it is uniformly decelerated, coming to a stop at B. The time taken for the whole journey is 5 minutes, and the magnitude of the acceleration is half that of the deceleration. Calculate the acceleration and the distance the car travels at constant speed.

Answer 1

$$s=v(T-t_1-t_2)+0.5at_1^2+0.5(2)at_2^2\\ 4000=\frac{80}{3.6}\left(5\cdot60-\frac{80}{3.6a}-\frac{80}{3.6(2)a}\right)\\+0.5a\left(\frac{80}{3.6a}\right)^2+0.5(2)a\left(\frac{80}{3.6(2)a}\right)^2\\a=0.14\frac{m}{s^2}$$

$$d=\frac{80}{3.6}\left(5\cdot60-\frac{80}{3.6(0.14)}-\frac{80}{3.6(2)(0.14)}\right)=1376\ m$$