Two particles A and B start from the same position along the same straight line. A goes with constant velocity 40m/s, whereas B stars from rest with constant acceleration 2ms-2. Then find the maximum distance between A and B before B overtakes A

Let at time t the maximum distance between A and B =S1-S2Therefore    S1-S2=40t-12×2×t2=40t-t2Now differentiating above eqn  with respect to time we get           dS1-S2dt=40-2tIn case  is maximum then dS1-S2dt=0Therefore          40-2t=0or                  t=20sThus maximum distance between A and B is              S1-S2=40×20-20×20=800-400=400m

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