Home
Class 11
PHYSICS
A point moves rectilinearly with deceler...

A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as `alpha=ksqrtv`, where k is a positive constant. At the initial moment the velocity of the point is equal to `V_0`. What distance will it take to cover that distance?

Text Solution

Verified by Experts

Let `t_(0)` be the time in which it comes to a stop.
Given that `-(dv)/(dt)=ksqrtV` `int_(0)^(t_(0))kdt=int_(v_(0))^(0)-(dv)/sqrtv`
`:. Kt_(0)=2sqrtv_(0) :. t_(0)=2/ksqrtv_(0)`
Now to find the distance covered before stopping,
`(dv)/(dt)=(dv)/(ds)(ds)/(dt)=v(dv)/(ds)` But ,`(dv)/(dt)=-ksqrtV,`
`:.v(dv)/(ds)=-ksqrtV ` `:. sqrtvdv=-kds`
`:.int_(v_(0))^(0)sqrtvdv=-int_(0)^(S)k dsrArr s =2/(3k)V_(0)^(3/2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the paticles as alpha=ksqrt(v) , where k is a positive constant .At the initial movement the velocity of the point in equal to V_(0) . What distance will it take to cover that distance?

If the velocity of a particle is v = At + Bt^2 , where A and B are constant, then the distance travelled by it between 1 s and 2 s is :

If the velocity of a particle is v = At + Bt^2 , where A and B are constant, then the distance travelled by it between 1 s and 2 s is :

A particle having a velocity v = v_0 at t= 0 is decelerated at the rate |a| = alpha sqrtv , where alpha is a positive constant.

A point moves with decleration along the circle of radius R so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment t=0 the velocity of the point equals v_0 . Find: (a) the velocity of the point as a function of time and as a function of the distance covered s_1 , (b) the total acceleration of the point as a function of velocity and the distance covered.

The velocity of a particle moving in the positive direction of the X-axis varies as V = KsqrtS where K is a positive constant. Draw V-t graph.

A particle moves in a circular path such that its speed v varies with distance as v=alphasqrts where alpha is a positive constant. Find the acceleration of particle after traversing a distance S?

A particle moves with an initial velocity V_(0) and retardation alpha v , where alpha is a constant and v is the velocity at any time t. Velocity of particle at time is :

The velocity of a particle moving in the positive direction of the x axis varies as v=alphasqrtx , where alpha is a positive constant. Assuming that at the moment t=0 the particle was located at the point x=0 , find: (a) the time dependence of the velocity and the acceleration of the particle, (b) the mean velocity of the particle averaged over the time that the particle takes to cover the first s metres of the path.

A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in moduli. At the initial moment t=0 the speed of the particle equals v_(0) , then th speed of the particle as a function of the distance covered S will be