Home
Class 12
MATHS
IF t=v^2/2, then (-(df)/(dt)) is equal t...

IF `t=v^2/2`, then `(-(df)/(dt))` is equal to ,
(where f is acceleration)

A

`f^2`

B

`f^3`

C

`-f^3`

D

`-f^2`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Similar Questions

Explore conceptually related problems

Let f be a differentiable function satisfying int_(0)^(f(x))f^(-1)(t)dt-int_(0)^(x)(cost-f(t)dt=0 and f((pi)/2)=2/(pi) The value of lim_(x to 0)(cosx)/(f(x)) is equal to where [.] denotes greatest integer function

A particle moves in such a way that its position vector at any time t is vec(r)=that(i)+1/2 t^(2)hat(j)+that(k) . Find as a function of time: (i) The velocity ((dvec(r))/(dt)) (ii) The speed (|(dvec(r))/(dt)|) (iii) The acceleration ((dvec(v))/(dt)) (iv) The magnitude of the acceleration (v) The magnitude of the component of acceleration along velocity (called tangential acceleration) (v) The magnitude of the component of acceleration perpendicular to velocity (called normal acceleration).

The position x of a particle varies with time t as x=at^(2)-bt^(3) . The acceleration at time t of the particle will be equal to zero, where (t) is equal to .

The position "x" of a particle varies with time "(t)" as "x=at^(2)-bt^(3)". The Acceleration at time t of the particle will be equal to zero, where "t" is equal to

If velocity of particle is given by v=2t^4 , then its acceleration ((dv)/(dt)) at any time t will be given by..

A body moves so that it follows the following relation (dv)/(dt)=-v^(2)+2v-1 where v is speed in m/s and t is time in second. If at t=0, v=0 then find the speed (in m/s) when acceleration in one fourth of its initial value.

If int_(0)^(x)f(t)dt=x^(2)+int_(x)^(1)(:t^(2)backslash f(t)dt, then f((1)/(2)) is equal to (a) (24)/(25) (b) (4)/(25) (c) (4)/(5) (d) (2)/(5)

f(x) = int_(x)^(x^(2))(e^(t))/(t)dt , then f'(t) is equal to :

A spaceship in space sweeps stationary interplanetary dust . As a result , its mass increase at a rate (dM(t))/(dt) =bv^(2) (t) , where v(t) is its instantaneous velocity . The instantaneous acceleration of the satellite is :