Question

A body is in motion along a straight line. As it crosses a fixed point a stop watch is started. The body travels a distance of 1.80m in the first 3 second and 2.20m in next 5 seconds what will be the velocity at the end of 9 seconds?

Answer 1

Let the initial velocity when the stop watch is started be u. Let the acceleration be a. In the first 3 s it travels a distance 1.80 m.

Using, S = ut + ½ at²

=> 1.8 = 3u + (9/2)a

=> 0.6 = u + 1.5a ………………(1)

And in (5+3=8) 8 s it travels (1.8+2.2=4) 4 m.

So,

4 = 5u + (25/2)a

=> 0.8 = u + 2.5a ……………(2)

From (1) and (2),

a = 0.2 m/s², u = 0.3 m/s

So, the velocity at the end of 9 s is,

v = u + at

=> v = 0.3 + (0.2)(9)

=> v = 0.3 + 1.8

=> v = 2.1 m/s