Answer to Question #145896 in Physics for yuvaraj

Question

A car starting from rest accelerates at rate α through a distance x then continues at constant speed for time t and then decelerates at a rate a/2​ to come to rest. If the total distance travelled is 15x. Then d =

Answer 1

$$v^2=2ax\to v=\sqrt{2ax}\\x_2=vt=\sqrt{2ax}t\\v^2=2(0.5)ax_3\to x_3=2x\\x+2x+\sqrt{2ax}t=15x\\x=\frac{at^2}{72}$$