Question

The displacement of a particle at rest ( t = 0) is given by $s = 6{t^2} - {t^3}.$ The time at which the particle will attain velocity equal to zero again is.
(A) 4 s B.) 8 s C.) 12 s D.) 16 s

Answer 1

Hint We are given relationship between ‘S’ i. e. displacement and time ‘t’ And the question given information about particle’s velocity, so me must know that we have differentiate ‘S’ w r t ‘t’:

Solution
So, $s = 6{t^2} + {t^3}$
Differentiate the above expression with respect to time , ‘t’
$ \Rightarrow \dfrac{{ds}}{{dy}} = v = 12t - 3{t^2}.$
This we get as the velocity and its relationship with time .
Now, we put $v = 0 \Rightarrow 12t - 3{t^2} = 0$
we get a quadratic equation in t , thus we find its roots by factorisation method by taking 3t common
$ \Rightarrow 3t\left( {4 - t} \right) = 0$
$ \Rightarrow t = 0,t = 4.$
So we have two time’s whose velocity is zero but in the question, we have an answer when “again” velocity is zero . $\therefore t = 4s$ is correct as t=0 is the time of starting when the object was already at rest.

NoteIdea of differentiation arises because we are given an expression of displacement dependent on the variable of time.Thus we differentiate it with respect to time . Similar is the case with velocity . On further differentiation of velocity we get the acceleration expression in terms of time . Do not make mistakes during differentiating the expressions and factorising the velocity expression to get ‘b’.