Home
Class 12
PHYSICS
A Particle move from the point (2.0 hat...

A Particle move from the point `(2.0 hati + 4.0 hatj)m`, at t 0 , with an initial velocity `(5.0 hati + 4.0hati)ms^(-1)`. It is acted upon by a constant force which prouduces a constant acceleration `(4.0hati + 4.0 hatj) ms^(-2)`. What is the distance of the particle from the origin at time 2 s?

A

15 m

B

`20sqrt(2)m`

C

5m

D

`10sqrt(2)m`

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • JEE MAINS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise Chemistry|1 Videos

  • JEE MAIN

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|452 Videos

  • JEE MAINS 2020

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise PHYSICS|250 Videos

Similar Questions

Explore conceptually related problems

A particle starts from origin at t = 0 with a velocity of 15 hati ms^(-1) and moves in xy-plane under the action of a force which produces a constant acceleration of 15 hati + 20 hatj ms^(-2) . Find the y-coordinate of the particle at the instant its x-coordinate is 180 m.

A particle starts from the origin at t=Os 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) . What time is the x -coordinate of the particle 16m ?

A particle P is at the origin starts with velocity u=(2hati-4hatj)m//s with constant acceleration (3hati+5hatj)m//s^(2) . After travelling for 2s its distance from the origin is

A particle starts from the origin at t=0 with an initial velocity of 3.0hati m/s and moves in the x-y plane with a constant cacceleration (6.0hati+4.0hatij)m//s^(2). The x-coordinate of the particle at the instant when its y-coordinates is 32 m is D meters. The value of D is :

Starting from the origin at time t = 0 , with initial velocity 5hatj "ms"^(-1) , a particle moves in the x - y plane with a constant acceleration of (10hati+4hatj)ms^(-2) . At time t , its coordinates are (20 m,y_0 m) . The value of t and y_0 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

The velocity of an object at t =0 is vecv_(0) =- 4 hatj m/s . It moves in plane with constant acceleration veca = ( 3hati + 8 hatj) m//s^(2) . What is its velocity after 1 s?

An object has a velocity, v = (2hati + 4hatj) ms^(-1) at time t = 0s . It undergoes a constant acceleration a = (hati - 3hatj)ms^(-2) for 4s. Then (i) Find the coordinates of the object if it is at origin at t = 0 (ii) Find the magnitude of its velocity at the end of 4s.

A particle moves from point (hati+hatj) m to a point (6hati-2hatj) m under the action of force (10hati-2hatj) N. The work done by the force is

A partical moves from a point (-2hati + 5hatj) to (4hati+ 3hatj) when a force of (4hati + 3hatj) N is applied . How much work has been done by the force?