Home
Class 12
PHYSICS
Velocity and acceleration of a particle ...

Velocity and acceleration of a particle at time `t=0`are `u=(2 hati+3 hatj) m//s and a=(4 hati+3 hatj) m//s^2` respectively. Find the velocity and displacement if particle at `t=2s.`

Text Solution

Verified by Experts

Here,
`veca = (4hati + 2hatj) m//s^(2)` is constant.
So, we can apply `vecv =vecu + vecat` and `vecs =vecu t + 1/2vecat^(2)`
Substuting the proper values, we get
`vecv =(2hati + 3hatj) + (2)(4hati + 2hatj) =(10hati + 7hatj) m//s`
Substuting the proper values, we get
`vecv=(2hati + 3hatj) +(2)(4hati + 2hatj) = (10hati + 7hatj)m//s`
and `vecx =(2)(2hati + 3hatj) +1/2(2)^(2) (4hati + 2hatj) = (12hati + 10hatj)m//s`
Therefore, velocity and displacement of particle at t=2s are `(10hati + 7hatj)`m/s and `(12hati + 10hatj)` m respectively.
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise SHORT ANSWER TYPE QUESTIONS|6 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise VERY SHORT ANSWER TYPE QUESTIONS|10 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (Circular Motion)|7 Videos

  • MOTION IN A STRAIGHT LINE

    AAKASH SERIES|Exercise very Short answer type question|15 Videos

Similar Questions

Explore conceptually related problems

Velocity and acceleration of a particle are v=(2 hati-4 hatj) m/s and a=(-2 hati+4 hatj) m/s^2 Which type of motion is this?

Velocity and acceleration of a particle at some instant of time are v = (3hati +4hatj) ms^(-1) and a =- (6hati +8hatj)ms^(-2) respectively. At the same instant particle is at origin. Maximum x-coordinate of particle will be

Velocity and acceleration of a particle are v=(2 hati) m/s and a = (4t hati+t^2 hatj) m /s^2 where, t is the time. Which type of motion is this ?

Acceleration of a particle at any time t is veca=(2thati+3t^(2)hatj)m//s^(2) .If initially particle is at rest, find the velocity of the particle at time t=2s

A particle is moving in x-y plane. Its initial velocity and acceleration are u=(4 hati+8 hatj) m//s and a=(2 hati-4 hatj) m//s^2. Find (a) the time when the particle will cross the x-axis. (b) x-coordinate of particle at this instant. (c) velocity of the particle at this instant. Initial coordinates of particle are (4m,10m).

Velocity of a particle at time t=0 is 2 hati m/s. A constant acceleration of 2 m/s^2 acts on the particle for 2 s at an angle of 60^@ with its initial velocity. Find the magnitude of velocity and displacement of particle at the end of t=2s.

The initial and final position vector for a particle are(-3m) hati+(2m)hatj+(8m)hatk and (9m)hati+(2m) hatj+(-8m)hatk respectively ,find the displacement of the particle.

Velocity of a particle at any time t is v=(2 hati+2t hatj) m//s. Find acceleration and displacement of particle at t=1s. Can we apply v=u+at or not?

Velocity of a particle in x-y plane at any time t is v=(2t hati+3t^2 hatj) m//s At t=0, particle starts from the co-ordinates (2m,4m). Find (a) acceleration of the particle at t=1s. (b) position vector and co-ordinates of the particle at t=2s.

Acceleration of a particle in x-y plane varies with time as a=(2t hati+3t^2 hatj) m//s^2 At time t =0, velocity of particle is 2 m//s along positive x direction and particle starts from origin. Find velocity and coordinates of particle at t=1s.