If `v=(t+2)(t+3)` then acceleration`(i.e(dv)/(dt))` at t=1 sec.

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

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

If displacement S at time t is S=t^(3)-3t^(2)-15t+12 , then acceleration at time t=1 sec is

If S=4 t^(3)-3t^(2)+2 , then acceleration is 42 "units" // sec ^(2) at the time t=

vec(r)=15t^(2)i+(20-20t^(2))j find magnitude of acceleration at t=1 sec.

vec(r)=15t^(2)i+(20-20t^(2))j find magnitude of acceleration at t=1 sec.

The angular displacement of a particle is given by theta =t^3 + t^2 + t +1 then,the angular acceleration of the particle at t=2 sec is ……. rad s^(-2)

The position of a particle is represented by the following equation. x=3t^(3)+7t^(2)+5t+8 where x is in metres and t in seconds. Find the acceleration of the particle at t = 1 s. Strategy : v=(dx)/(dt) and a=(dv)/(dt)

The position of a particle is represented by the following equation. x=3t^(3)+7t^(2)+5t+8 where x is in metres and t in seconds. Find the acceleration of the particle at t = 1 s. Strategy : v=(dx)/(dt) and a=(dv)/(dt)

If v=(t^(2)-4t+10^(5)) m/s where t is in second. Find acceleration at t=1 sec.