Question

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

Answer 1

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

Distance = velocity x time
BC = 10 x 15.5
= 155 m