The position vector of a particle changes with time according to the relation `vec(r ) (t) = 15 t^(2) hat(i) + (4 - 20 t^(2)) hat(j)` What is the magnitude of the acceleration at t = 1?

The position vector of a particle changes with time according to the direction vec(r)(t)=15 t^(2)hati+(4-20 t^(2))hatj . What is the magnitude of the acceleration at t=1?

If the velocity is vec(v) = 2 hat(i) + t^(2) hat(j) - 9 vec(k) , then the magnitude of acceleration at t = 0 . 5 s is :

If the position vector of a particle is given by vec(r ) = (cos 2t) hat(i) + (sin 2 t) hat(j) + 6 t hat(k) m . Calculate magnitude of its acceleration (in m//s^(2) ) at t = (pi)/(4)

Position vector vec(r) of a particle varies with time t accordin to the law vec(r)=(1/2 t^(2))hat(i)-(4/3t^(1.5))hat(j)+(2t)hat(k) , where r is in meters and t is in seconds. (a) Find suitable expression for its velocity and acceleration as function of time (b) Find magnitude of its displacement and distance traveled in the time interval t=0 to 4 s .

Position vector vec(r) of a particle varies with time t accordin to the law vec(r)=(1/2 t^(2))hat(i)-(4/3t^(1.5))hat(j)+(2t)hat(k) , where r is in meters and t is in seconds. Find suitable expression for its velocity and acceleration as function of time.

Position vector vec(r) of a particle varies with time t accordin to the law vec(r)=(1/2 t^(2))hat(i)-(4/3t^(3))hat(j)+(2t)hat(k) , where r is in meters and t is in seconds. Find suitable expression for its velocity and acceleration as function of time

The position vector vec(r) of a moving particle at time t after the start of the motion is given by vec(r) = (5+20t)hat(i) + (95 + 10t - 5 t^(2)) hat(j) . At the t = T, the particle is moving at right angles to its initial direction of motion. Find the value of T and the distance of the particle from its initial position at this time.

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.