A car starting from rest accelerates at constant rate f through a distance S, then continues at constant speed for time t and then decelerates at rate f/2 to rest. If the total distance traversed is 15S, then S equals:
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Question

A car starting from rest accelerates at constant rate f through a distance S, then continues at constant speed for time t and then decelerates at rate f/2 to rest. If the total distance traversed is 15S, then S equals:

Options

G1 Physics · Accepted Answer

After accelerating through S from rest, v²=2fS. Stopping with retardation f/2 requires distance v²/f=2S. Hence constant-speed distance =15S-S-2S=12S=vt. Using v=12S/t in v²=2fS gives S=ft²/72. Hence option C.