Question

a body moving with constant acceleration covers the distance between two points 60m apart in 5 sec. Its velocity as it passes the 2nd point is 15m/sec. What is the magnitude of average velocity

Answer 1

Let’s review the 4 fundamental kinematic equations of motion for constant acceleration (strongly recommend you commit these to memory – they will serve you well):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 5m/s, s = 30m, t = 3s

So we find a from (1) s = ut + ½at^2

30 = 5(3) + ½a(9)

So a = 15/4.5 = 3.333m/s^2

Then v = u + at = 5 + 3.333(3) = 15m/s

And this is used as initial velocity in s = ut + ½at^2 where t = 2s

s = 15(2) + ½(3.333)(4)

s = 30 + 6.667 = 36.667

The distance traveled in the next 2s is 36.667m