## Question

### Solution

Correct option is

– 9 m/s

As according to given problem,

S = t3 – 6t+ 3+ 4

Instantaneous velocity

v = ds/dt = 3t2 – 12+ 3                        …(1)

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

v = 3 × 22 – 12 × 2 + 3 = – 9 m/s.

#### SIMILAR QUESTIONS

Q1

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.

Q2

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?

Q3

A particle moving with velocity equal to 0.4 m/s is subjected to an acceleration of  0.15 m/s2 for 2 sec in a direction at right angle to its direction of motion. What is the magnitude of resultant velocity?

Q4

Two particles A and B move with constant velocities v1 and v2 along two mutually perpendicular straight lines towards the intersection point O. At moment t = 0 the particle were located at distances l1 and l2 from O respectively. Find the time when they are nearest and also this shortest distance.