Home
Class 11
PHYSICS
An object , moving with a speed of 6.25...

An object , moving with a speed of ` 6.25 m//s `, is decelerated at a rate given by :
`(dv)/(dt) = - 2.5 sqrt (v)` where `v` is the instantaneous speed . The time taken by the object , to come to rest , would be :

A

`4s`

B

`8s`

C

`1s`

D

`2s`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

An object moving with a speed of 6.25 m//s , is deceleration at a rate given by : (dv)/(dt) = -25 sqrt(v) , where v is instantaneous speed. The time taken by the obeject, to come to rest, would be :

An object moving with a speed of 6.25m/s , is decreased at a rate given by (dv/dt)=-2.5sqrt(v) where v is instantaneous speed, The time taken by the object to come to rest, would be

An object is subjected to retardation, (dv)/(dt)=-5sqrtv , which has initial velocity of 4ms^(-1) . The time taken by the object to come to rest would be

An object of 10 kg is moving with a speed of 2 m/s. The kinetic energy of the object is:

A particle initially (i.e., at t = 0) moving with a velocity u is subjected to a retarding force, as a result of which it decelerates a=-ksqrt v at a rate where v is the instantaneous velocity and k is a positive constant. The time T taken by the particle to come to rest is given by :

An object is moving with a speed 100 m/s. Find the distance travelled by this object in one minute

A point moves linearly with deceleration which is given by dv//dt=-alphasqrt(v) , where alpha is a positive constant. At the start v=v_(0) . The distance traveled by particle before it stops will be

The speed(v) of a particle moving along a straight line is given by v=(t^(2)+3t-4 where v is in m/s and t in seconds. Find time t at which the particle will momentarily come to rest.