Home
Class 12
MATHS
Displacement s of a particle at time t i...

Displacement `s` of a particle at time `t` is expressed as `s=1/2t^3-6t`. Find the acceleration at the time when the velocity vanishes (i.e., velocity tends to zero).

Text Solution

Verified by Experts

The correct Answer is:
a=3t=6
Promotional Banner

Topper's Solved these Questions

  • DY / DX AS A RATE MEASURER AND TANGENTS, NORMALS

    ARIHANT MATHS|Exercise EXAMPLE|6 Videos
  • DY / DX AS A RATE MEASURER AND TANGENTS, NORMALS

    ARIHANT MATHS|Exercise SINGLE OPTION CORRECT TYPE QUESTIONS|10 Videos
  • DIFFERENTIATION

    ARIHANT MATHS|Exercise Exercise For Session 10|4 Videos
  • ELLIPSE

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|26 Videos

Similar Questions

Explore conceptually related problems

If the displacement of a particle is givne by s=((1)/(2)t^(2)+4sqrtt)m . Find the velocity and acceleration at t = 4 seconds.

The displacement of a particle along a straight time at line t is given by ,x=4+2t +3t^2+4t^3 .Find acceleration fo the particle at t=2 second.

The displacement of a particle is directly proportional to the square of the time.Explain whether the object is moving with uniform velocity or with uniform acceleration.

The displacement of a particle is SHM is x=10 simn(2t-pi/6) meter.When its displacement is 6m,the velocit of the particle (in m s^(-1)) is :

A point moves in a straight line and after t seconds, its accelearation is (2t+1)(cm)/sec^2 . If its velocity is 4cm/sec when t=0, find its velocity after 2 seconds and the distance moved it by in 3 seconds.

RMS velocity of a particle is v_(r.m.s.) at pressure P. If pressure is increased by two times, the r.m.s. velocity becomes:

The acceleration fo a body in ms^-2 starting from rest varies with time in secnds as per the relation a = 3t + 4 then the magnitude of velocity of the body at t=2 s will be :

If the mass of a planet is 2 times that of earth and radius is 3 times that of earth ,then find the escape velocity for that planet.For earth ,escape velocity =11.2 km s^(-1)

A particle is moving in a straight line such that its displacement s at an instant of time t is given by the relation s = 3t^2 + 4 t + 5 Find the initial velocity of the particle.