Need help with calculus problem

Question

A particle starts from rest and moves in a straight line with constant acceleration passing through 2 points A and B, which are 100m apart in 15s.If the velocity of the particle at B is three times its velocity at A, find the velocity of the particle at A and its acceleration

Answer 1

Hi William,

 

distance = 1.2at^2

Since, acceleration is constant,

100 = 1/2*15^2a

a = 8/9 ms^-2

Let the time it takes for it to get to point A from the start be x. Then. we can write below based on the given relationship between the velocities

8/9x = v

8/9(x+15) = 3v

3*(8/9x) = 8/9(x+15)

x = 15/2 sec

Now, we can find the velocity at point A.

V_A = (15/2)*(8/9)

= 20/3 ms^-1

The acceleration at A is 8/9 ms^-2