## Question

A particle moves along a straight line such that its displacement at any time*t* is given by s = (*t ^{3}* + 6

*t*

^{2}+ 3

*t*+ 4)m. What is the velocity of the particle when its acceleration is zero?

### Solution

– 9 *m/s*

As according to given problem,

S = *t*^{3} – 6*t*^{2 }+ 3*t *+ 4

Instantaneous velocity

*v =* *ds/dt* = 3*t*^{2} – 12*t *+ 3 …(1)

and acceleration

So acceleration will be zero when 6*t* – 12 = 0, i.e., *t* = 2 sec. And the so velocity when acceleration is zero, *i.e.,* at *t * = 2 sec from Eqn. (1), will be

*v* = 3 × 2^{2} – 12 × 2 + 3 = – 9 *m/s.*

#### SIMILAR QUESTIONS

If the initial velocity of the particle is u and collinear acceleration at any time *t* is at, calculate the velocity of the particle after time *t*.

A particle starts moving from the position of rest under a constant acc. If it travels a distance *x* in *t* sec, what distance will it travel in next *t* sec?

A particle moving with velocity equal to 0.4 *m/s* is subjected to an acceleration of* * 0.15 *m/s*^{2} for 2 sec in a direction at right angle to its direction of motion. What is the magnitude of resultant velocity?

Two particles *A* and *B* move with constant velocities *v*_{1} and *v*_{2} along two mutually perpendicular straight lines towards the intersection point *O*. At moment *t* = 0 the particle were located at distances *l*_{1} and *l*_{2} from O respectively. Find the time when they are nearest and also this shortest distance.