Home
Class 12
PHYSICS
The velocity of any particle is related ...

The velocity of any particle is related with its displacement As, `x = sqrt(v+1)`, Calculate acceleration at `x = 5m`.

A

`sqrt(6) ` unit

B

24 unit

C

240 unit

D

25 unit

Text Solution

Verified by Experts

` x = sqrt(v + 1 ) rArr x^(2) + v + 1 rArr v = x^(2) - 1 `
` rArr (dv)/(dx) = 2x ` . Now acceleration
` a = v (dv)/(dx) = (x^(2) - 1) . 2x `
This is the acceleration at position x. Now at x = 5 unit, ` a= (5^(2) -1) (2xx5) = 240` unit
Hence correct answer is (C).
Promotional Banner

Topper's Solved these Questions

  • ONE DIMENSION MOTION

    MOTION|Exercise EXERCSE -1 (Section - A : Distance, Displacement, Velocity and Acceleration, Equation of Motion )|26 Videos

  • ONE DIMENSION MOTION

    MOTION|Exercise EXERCSE -1 (Section - B : Motion under Gravity)|11 Videos

  • NLM & FRICTION

    MOTION|Exercise EXERCISE-4 ( LEVEL-II)|15 Videos

  • OPTICS

    MOTION|Exercise Exercise|45 Videos

Similar Questions

Explore conceptually related problems

" If the velocity v of a particle varies as the square of its displacement "x" then the acceleration varies as "

If the velocity "v" of a particle varies as the square of its displacement "x" then the acceleration varies as

The velocity of a particle varies with its displacement as v=(sqrt(9-x^2))ms^-1 . Find the magnitude of the maximum acceleration of the particle.

If the velocity of a body related to displacement x is given by v=sqrt(5000+24x) m/s, then the acceleration of the body is _______ m//s^(2) .

The velocity v of moving object varies with displacement u as v=3x^(2)-2 ,then the acceleration at x=2m is

If v = sqrt(5000+24x) m/s then calculate acceleration.

Velocity of a particle moving in a straight line varies with its displacement as v = sqrt(4+4x)m//s where x is displacement. Displacement of particle at time t = 0 is x = 0. Find displacement (in m) of particle in 2 sec.

The velocity of particle is given by v=sqrt(gx) . Find its acceleration.

The velocity time relation of a particle is given by v = (3t^(2) -2t-1) m//s Calculate the position and acceleration of the particle when velocity of the particle is zero . Given the initial position of the particle is 5m .

If the velocity v of particle moving along a straight line decreases linearly with its displacement s from 20 ms^(-1) to a value approaching zero at s = 30 m, then acceleration of the particle at v=10ms^(-1) is