Position of a body with acceleration `a` is given by `x=Ka^mt^n`, here t is time Find demension of m and n.

Position of a body with acceletation a is given box x=ka^(m)t^(n) , where t is time and k is numeric constant. Find the value of m.

Completion type Question : The position of a body with respect to time is given by x = 2t^(3) - 6t^(2) + 92t + 6 . If x is in metres and t in seconds, then acceleration at t = 0 is

The position of a body with respect to time is given by x = 4t^(3) – 6t^(2) + 20 t + 12 . Acceleration at t = 0 will be-

If position of a particle at instant t is given by x=3t^(2) , find the velocity and acceleration of the particle.

The position of a particle moving in a straight line is given by x=3t^(3)-18t^(2)+36t Here, x is in m and t in second. Then

The position of a particle is given by the equation f(t)=t^(3)-6t^(2)+9t where t is measured in second and s in meter. Find the acceleration at time t. What is the acceleration at 4 s?

The position of a particle is given by the equation f(t)=t^(3)-6t^(2)+9t where t is measured in second and s in meter. Find the acceleration at time t. What is the acceleration at 4 s?

A particle moves such that its acceleration is given by a =- beta (x-2) Here beta is positive constant and x is the position form origin. Time period of oscillation is

A particle moves such that its acceleration is given by a = -Beta(x-2) Here, Beta is a positive constnt and x the x-coordinate with respect to the origin. Time period of oscillation is