Home
Class 12
MATHS
The distance covered by a particle movin...

The distance covered by a particle moving in a straight line from a fixed point on the line is `s ,` where `s^2=a t^2+2b t+cdot` Then prove that acceleration is proportional to `s^(-3)dot`

Promotional Banner

Topper's Solved these Questions

  • APPLICATION OF DERIVATIVES

    CENGAGE|Exercise Exercise 5.6|5 Videos
  • APPLICATION OF DERIVATIVES

    CENGAGE|Exercise Exercise 5.7|8 Videos
  • APPLICATION OF DERIVATIVES

    CENGAGE|Exercise Exercise 5.4|6 Videos
  • 3D COORDINATION SYSTEM

    CENGAGE|Exercise DPP 3.1|11 Videos
  • APPLICATION OF INTEGRALS

    CENGAGE|Exercise Solved Examples And Exercises|137 Videos

Similar Questions

Explore conceptually related problems

The distance covered by a particle moving in a straight line from a fixed point on the line is s where s^(2)=at^(2)+2bt+c .Then prove that acceleration is proportional to s^(-3) .

If the velocity v of a particle moving along a straight line and its distance s from a fixed point on the line are related by v^(2)=a^(2)+s^(2) , then its acceleration equals

If a particle moving in a straight line and its distance x cms from a fixed point O on the line is given by x=sqrt(1+t^(2)) cms, then acceleration of the particle at t sec. is

The distance traversed by a particle moving along a straight line is given by x = 180 t + 50 t^(2) metre. Find the velocity at the end of 4s.

A particle is moving on a straight line and its distance x cms from a fixed point O on the line is given by x=sqrt(t^(2)+1) then the velocity of particle at t=1 is

A particle moves in a straight line so that s=sqrt(t) , then its acceleration is proportional to

The distance s travelled by a particle moving on a straight line in time t sec is given by s=2t^(3)-9t^(2)+12t+6 then the initial velocity of the particle is