Home
Class 12
PHYSICS
A particle leaves the origin with an ini...

A particle leaves the origin with an initial velodty `v= (3.00 hati) m//s` and a constant acceleration `a= (-1.00 hati-0.500 hatj) m//s^2.` When the particle reaches its maximum x coordinate, what are
(a) its velocity and (b) its position vector?

A

`-2.0 ms^(-1)`

B

`-1.0 ms^(-1)`

C

`-1.5 ms^(-1)`

D

`-1.0 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C

At maximum x coordinate `v_(x)=0, 3-t=0 rArr t=3 sec, v_(x)=0 +(-0.5) (3)=-1.5 m//s`.
Promotional Banner

Similar Questions

Explore conceptually related problems

From the origin, a particle starts at t = 0 s with a velocity vecv=7.0hatim//s and moves in the xy plane with a constant acceleration of veca=(-9.0hati+3.0hatj)m//s^(2) . At the time the particle reaches the maximum x coordinate, what is its (a) velocity and (b) position vector?

A particle leaves the origin with an initial velocity vecv=3hati m/s and a constant acceleration veca=(-hati-0.5hatj) m//s^(2) in free space. Its velocity vecv and position vector vecr when it reaches its maximum x-coordinate are

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 :

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=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 has an initial velocity (6hati+8hatj) ms^(-1) and an acceleration of (0.8hati+0.6hatj)ms^(-2) . Its speed after 10s 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 has an initial velocity of 4 hati +3 hatj and an acceleration of 0.4 hati + 0.3 hatj . Its speed after 10s is

A particle is given an initial velocity of vecu=(3 hati+4 hatj) m//s . Acceleration of the particle is veca=(3t^(2) +2 thatj) m//s^(2) . Find the velocity of particle at t=2s.

A particle at origin (0,0) moving with initial velocity u = 5 m/s hatj and acceleration 10 hati + 4 hatj . After t time it reaches at position (20,y) then find t & y