Position vector of a particle moving in space is given by :
`vec(r)=3sin t hat i+3 cos t hatj+4 t hatk`
Distance travelled by the particle in `2s` is :

(a) A particle starts moving at t = 0 in x-y plane such that its coordinates (in cm) with time (in sec) change as x = 3 t and y = 4 sin (3t) . Draw the path of the particle. (b) If position vector of a particle is given by vec(r) = (4t^(2) - 16t) hati +(3t^(2) - 12t) hatj , then find distance travelled in first 4 sec.

Position of a particle at any instant t is given by vec r=3t hat i + 2t^2 haij + 5 hat k . Its velocity at same instant will be

If the position vector of a particle is given by vec r= 5t^(2)hat i +7t hat j +4 hat k , then its velocity lies in:

A particle is moving with a position vector, vec(r)=[a_(0) sin (2pi t) hat(i)+a_(0) cos (2pi t) hat(j)] . find Distance travelled by the particle in 1 sec is

If the position vector of the particle is given by vecr = 3 t^2 hati + 5t hatj + 4hatk . Find the velocity of the particle at t = 3s.

Position vector of a particle moving in x-y plane at time t is r=a(1- cos omega t)hat(i)+a sin omega t hat(j) . The path of the particle is

Position vector of a particle moving in x-y plane at time t is r=a(1- cos omega t)hat(i)+a sin omega t hat(j) . The path of the particle is

The position vector of a particle is given by vecr=(2 sin 2t)hati+(3+ cos 2t)hatj+(8t)hatk . Determine its velocity and acceleration at t=pi//3 .