A car, starting from rest, accelerates at a rate of through a distance S, then continues at constant speed for time t and then decelerates at a rate of 2 to come to rest. If the total distance travelled is 10 S , then ratio S t^2 is

Step-by-step solution

Let the maximum speed reached by the car be v . For the first part of the motion, the car accelerates from rest at over a distance S . Using the equation of kinematics: v^2 = 0 + 2 S v = 2 S For the second part of the motion, the car travels at a constant speed v for time t . The distance covered is: S₂ = vt For the third part of the motion, the car decelerates at 2 to come to rest. The distance covered S₃ is given by: 0 = v^2 - 2 ( 2 )S₃ S₃ = v^2 Substituting v^2 = 2 S gives: S₃ = 2 S = 2S The total distance trave