Question

A particle is moving with uniform acceleration along a straight line passes three successive points a,b and c where the distance ab:bc is as 3:5 and the time taken from a to b is 40 seconds . If the velocities at a and c are 5 m/s and 15m/s , find the velocity of the particle at b

Answer 1

Answer:

Velocity will be 20m/s.

Explanation:

If we let the velocity as the point b is x m/s and we know from the question that ab:bc is given as 3:5. So, let the distance be y so they will be 3y and 5y or we can take the distance to be 3+5 = 8m.

Now, applying Newtons laws of motion we will get v^2 = u^2 +2aS on substituting the values of v u and S we will get that 900 = 100 + 2*a*8 or on solving we will get the value of acceleration a to be 50m/s^2.

So, velocity at b will be x^2 = 100 + 2*50*3 so on solving we will get the value of x to be 20m/s.