Two particles are moving along x-axis. Particle-1 is `40 m` behind Particle-2. Particle-1 starts with velocity`12 m//s` and acceleration `4 m//s^2` both in positive x-direction. Particle-2 starts with velocity `4 m//s` and acceleration `12 m//s^2` also in positive x-direction. Find
(a) the time when distance between them is minimum.
(b) the minimum distacne between them.

Two particles are moving along x-axis. Particle-1 starts from x=-10 m with velocity 4 m//s along negative x-direction and acceleration 2 m//s^2 along positive x-direction. Particle-2 starts from x=+2 m with velocity 6 m//s along positive x-direction and acceleration 2 m//s^2 along negative x-direction. (a) Find the time when they collide. (b) Find the x-coordinates where they collide. Both start simultaneously.

Two particles are moving along x-axis. Particle-1 starts from x=-10 m with velocity 4 m//s along negative x-direction and acceleration 2 m//s^2 along positive x-direction. Particle-2 starts from x=+2 m with velocity 6 m//s along positive x-direction and acceleration 2 m//s^2 along negative x-direction. (a) Find the time when they collide. (b) Find the x-coordinates where they collide. Both start simultaneously.

A particle is moving with initial velocity 1m/s and acceleration a=(4t+3)m/s^(2) . Find velocity of particle at t=2sec .

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?

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 ?

A particle starts with an initial velocity 2.5 m//s along the posiive x-direction and it accelerates uniformly at the rate 0.50 m//s^2. Find the distance travelled by it in the first two seconds

To particles P and Q are initially 40 m apart P behind Q. Particle P starts moving with a uniform velocity 10 m/s towards Q. Particle Q starting from rest has an acceleration 2m//s^(2) in the direction of velocity of P. Then the minimum distance between P and Q will be

A particle starts from rest and moves with an acceleration of a={2+|t-2|}m//s^(2) The velocity of the particle at t=4 sec is

A particle starts from rest and moves with an acceleration of a={2+|t-2|}m//s^(2) The velocity of the particle at t=4 sec is

A particle starts from rest and moves with an acceleration of a={2+|t-2|}m//s^(2) The velocity of the particle at t=4 sec is