If the velocity of a particle is (10+2t^(2)m/s ,then the average acceleration of the particle between 2s and 5s is (in m/s^2)

. If the velocity of a particle is (10 + 2 t 2) m/s , then the average acceleration of the particle between 2s and 5s is

If velocity of a particle is given by v=2t^(2)-2 then find the acceleration of particle at t = 2 s.

Velocity of a particle is given as v = (2t^(2) - 3)m//s . The acceleration of particle at t = 3s will be :

The position of a particle moving on the x-axis is given by x=t^(3)+4t^(2)-2t+5 where x is in meter and t is in seconds (i) the velocity of the particle at t=4 s is 78 m//s (ii) the acceleration of the particle at t=4 s is 32 m//s^(2) (iii) the average velocity during the interval t=0 to t=4 s is 30 m//s (iv) the average acceleration during the interval t=0 to t=4 s is 20 m//s^(2)

The body moves along the x- axis . At time t=0 the velocity of particle is +0.5m//s . The acceleration of particle in different time interval is given as : a={{:(-1 m//s^(2)," ",0slt=tlt=3s),(-2 m//s^(2)," ",3s lt tlt6s),(+4m//s^(2)," ",6s lt=tlt=10s):}

If velocity of a particle is given by v=(2t+3)m//s . Then average velocity in interval 0letle1 s is :

A particle is moving under a constant acceleration of 3m//s^(2) . The initial velocity of the particle is 10m/s. calculate the distance travelled by the particle is 5s and also in th 5th second of its motion.

The velocity time relation of a particle is given by v = (3t^(2) -2t-1) m//s Calculate the position and acceleration of the particle when velocity of the particle is zero . Given the initial position of the particle is 5m .

A particle experience an acceleration a=(4e^(-3t))m/s^(2) along positive x-axis, where t is time in s .If the initial velocity of the particle is 2m/s along positive x-axis,the velocity of the particle when its acceleration becomes zero is?