Home
Class 11
PHYSICS
Velocity and acceleration of a particle ...

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?

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
v and a both are constant vectors. Further, these two
vectors are antiparallel.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • KINEMATICS

    DC PANDEY ENGLISH|Exercise Exercise 6.3|6 Videos

  • KINEMATICS

    DC PANDEY ENGLISH|Exercise Exercise 6.4|2 Videos

  • KINEMATICS

    DC PANDEY ENGLISH|Exercise Exercise 6.1|2 Videos

  • GRAVITATION

    DC PANDEY ENGLISH|Exercise (C) Chapter Exercises|45 Videos

  • KINEMATICS 1

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|15 Videos

Similar Questions

Explore conceptually related problems

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 ?

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.

Assertion : Velocity and acceleration of a particle are given as, v=hati-hatj and a=-2 hati+2 hatj This is a two dimensional motion with constant acceleration. Reason : Velocity and acceleration are two constant vectors.

velocity and acceleration of a particle at some instant are v=(3 hati-4 hatj+2 hatk) m//s and a=(2 hati+hatj-2hatk) m//s^2 (a) What is the value of dot product of v and a at the given instant? (b) What is the angle between v and a, acute, obtuse or 90^@ ? (c) At the given instant, whether speed of the particle is increasing, decreasing or constant?

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 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

A bomb initially at rest explodes by it self into three equal mass fragments. The velocities of two fragments are ( 3 hati + 2 hatj ) m/s and (–hati – 4 hatj ) m/s. The velocity of the third fragment is (in m/s)-

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.

Velocity of a particle changes from (3hati +4hatj)m//s to (6hati +5hatj)m//s 2s. Find magnitude and direction of average acceleration.

A particle is moving in such a way that at one instant velocity vector of the particle is 3hati+4hatj m//s and acceleration vector is -25hati - 25hatj m//s ? The radius of curvature of the trajectory of the particle at that instant is

Home
Profile