A particle P, starting from rest at A, moves in a straight line ABCD. It accelerates uniformly at 5ms-2 from A to B. From B to C
it travels at constant velocity, and from C to D it slows down uniformly at 4ms-2, coming to rest at D. Given that AB=10m and that the total time P the particle is in motion is 20s, find the distance BC
First, we find the time spent accelerating: s = ut + 1/2 at²; u = 0 10 = 1/2 5t² t = 2 seconds Velocity at point B: v = u + at; u = 0, a = 5 m/s², t = 2 s v = (5)(2) = 10 m/s
Now, the same velocity is at C so we find the time decelerating to 0 in CD v = u + at, v = 0, u = 10 m/s, a = -4 m/s² 0 = 10 - 4t t = 2.5 seconds Total time = time(AB) + time(BC) + time(CD) 20 = 2 + time(BC) + 2.5 time(BC) = 15.5 seconds
