A particle is moving in a plane according to the law `x=at^2`,`y=bt^3` where `a=2m/s^2`, `b=1m/s^3`. Radius of curvature of particle at t=1s is found to be `eta^3/12m`. Find `eta`.

A particle moves along x-axis according to the law x=(t^(3)-3t^(2)-9t+5)m . Then :-

A particle moves in the x-y plane according to the law x=t^(2) , y = 2t. Find: (a) velocity and acceleration of the particle as a function of time, (b) the speed and rate of change of speed of the particle as a function of time, (c) the distance travelled by the particle as a function of time. (d) the radius of curvature of the particle as a function of time.

A particle moves along a straight line according to the law s=16-2t+3t^(3) , where s metres is the distance of the particle from a fixed point at the end of t second. The acceleration of the particle at the end of 2s is

A particle starts moving with velocity "vecV=hat i(m/s)", and acceleration "vec a=(hat i+hat j)(m/s^(2))".Find the radius of curvature of the particle at "t=1sec".

A particle is moving in a circle of radius 1 m with speed varying with time as v = (2t) m/s. In first 2 sec:

A particle is moving in a circle of radius 1 m with speed varying with time as v=(2t)m//s . In first 2 s

A particle is moving on a circular path of radius 1 m with 2 m/s. If speed starts increasing at a rate of 2m//s^(2) , then acceleration of particle is

A particle moves according to law s=t^(3)-6t^(2)+9t+15 . Find the velocity when t = 0.