A particle P movies in straight line such that, t s after leaving a Point O, its velocity `vms^-1`
￼is given by `v=36t-3t^2`
￼ for t≥ 0
find the distance of P from O when P is at instantaneous rest?
At instantaneous rest, velocity, v = 0
i.e. 36t - 3t^2 = 0
or, 3t (12-t) = 0
There are two solutions to this, t = 0 sec and t = 12 sec.
We will discount the scenario of t = 0 sec, since at that time particle left point O, distance =0, i.e. motion has just started.
You can also try and check the values of v for t=11 sec and t=13 sec to see that at 11 sec, velocity is positive and at 13 seconds, it is negative; to confirm that the particle is at instantaneous rest at t = 12 sec.
For t = 12 sec,
The distance traveled by particle P from O is given by
Distance = `int_0^12 vdt = int_0^12 (36t-3t^2)dt = 36t^2/2 - 3t^3/3 = 18t^2 - t^3`
for limits between 0 to 12 seconds.
Thus, distance = 18 x 12^2 - 12^3 - 18x0^2 + 0^3 = 2592 - 1728 = 864 m.
Hope this helps.