Home
Class 11
PHYSICS
A particle moves through the origin of a...

A particle moves through the origin of an `xy-` cordinate system at `t=0` with initial velocity `u=4i-5jm//s`. The particle moves in the `xy` plane with an acceleration `a=2im//s^(2)`. Spped of the particle at `t=4` second is `:`

Promotional Banner

Similar Questions

Explore conceptually related problems

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

A particle starts from the origin at t= 0 s with a velocity of 10.0 hatj m//s and moves in the xy -plane with a constant acceleration of (8hati+2hatj)m//s^(-2) . Then y -coordinate of the particle in 2 sec is

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

Starting from the origin at time t=0 ,with initial velocity [5hat jms^(-1) ,a particle moves in the x -y plane with a constant acceleration of ( 10ˆi+4ˆj)m/s 2 ms^(-2).At time t ,its coordinates are (20m,y_(0)m are,respectively:

A particle starts from the origin at t = 0 s with a velocity of 10.0 hatj m/s and moves in plane with a constant acceleration of (8hati + 2hatj)ms^(-2) . The y-coordinate of the particle in 2 sec is

A particle starts from the origin at t = 0 s with a velocity of 10.0 hatj m/s and moves in plane with a constant acceleration of (8hati + 2hatj)ms^(-2) . The y-coordinate of the particle in 2 sec is