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

Let the three time periods be x,y,z seconds

x+y+z = 20
(5/2 x^2) + (5xy) + (5xz-2z^2) = 20

Now, since it decelerates at 4/5 the rate that it took to accelerate, it will take 5/4 as long to come to rest. So, our third equation is

z = 5/4 x

Now just solve for the time periods, and then you can get the three distances.