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

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

A body moving with some initial velocity and having uniform acceleration attains a final velocity v m/s after travelling x m. If its final velocity is v = sqrt (180 - 7x) , find the acceleration of the body.

A wheel of radius R=1 m rolls on ground with uniform velocity v=2 m/s . Calculate the relative acceleration of topmost point of wheel with respect to bottom most point (in m//s^(2) ).

A particle of mass m moves in a circle of radius R in such a way that its speed (v) varies with distance (s) as v = a sqrt(s) where a is a constant. Calcualte the acceleration and force on the particle.

If the velocity of a particle is given by v=(180-16x)^((1)/(2))(m)/(s) , then its acceleration will be

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 .

The motion of a body is given by the equation dv//dt=6-3v , where v is in m//s. If the body was at rest at t=0 (i) the terminal speed is 2 m//s (ii) the magnitude of the initial acceleration is 6 m//s^(2) (iii) The speed varies with time as v=2(1-e^(-3t)) m//s (iv) The speed is 1 m//s , when the acceleration is half initial value

The diagram shows variation of (1)/(v) with respect to time (where v is in m/s). What is the instantaneous acceleration of body at t=3 sec. (in (m)/(s^(2)) ).

For motion of an object along the x-axis the velocity v dipends on the displacement x an v=3x^(2) , then what is the acceleration at x=2 m .