COMEDK202510 May 2025Evening ShiftPhysicsMotion in One DimensionActual
A car, starting from rest, accelerates at the rate (f) through a distance (S), then continues at constant speed for some time ( t ) and then decelerates at the rate f 2 to come to rest. If the total distance is 5 S , then
Options
- AS= 1 2 f t^2
- BS=4 f t^2
- CS= 1 4 f t^2
- DS=2 f t^2
Correct answer
A. S= 1 2 f t^2
Step-by-step solution
Let the car accelerate from rest at rate f for distance S . The final velocity v reached is given by v^2 = 0^2 + 2fS , so v = 2fS . The time taken for this phase is t₁ = v f = 2fS f = 2S f . The car then travels at constant speed v for time t . The distance covered in this phase is S₂ = vt = t 2fS . Finally, the car decelerates at rate f 2 to come to rest. Let the distance covered be S₃ . Using v^2 = 2aS₃ , we have v^2 = 2( f 2 )S₃ , which gives S₃ = v^2 f = 2fS f = 2S . The total distance is given as 5S . Thus, S