Home
Class 11
PHYSICS
A particle starts moving from point (2, ...

A particle starts moving from point (2, 10, 1). Displacement for the particle is `8hati - 2hatj + hatk` . The final coordinates of the particle is

A

`(10, 8, 2)`

B

`(8, 10, 2)`

C

`(2, 10, 8)`

D

`(8, 2, 10)`

Text Solution

Verified by Experts

The correct Answer is:
A
Initial position vector of the particle, `vecr_(i) = 2hati + 10 hatj + hatk`
Let final position vector of the particle be
`vecr_(f) = x hati + y hatj + zhatk`
`therefore` Displacement, `trianglevecr = vecr_(f) - vecr_(i)`
` = 8hati - 2hatj + hatk = x hati + y hatj + zhatk - (2hati + 10 hatj + hatk)`
or `xhati + yhatj + zhatk = 10 hati + 8hatj + 2hatk`
Hence, the final coordinates of the particle are `(`10, 8, 2)`
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    NCERT FINGERTIPS|Exercise Motion In A Plane With A Constant Acceleration|2 Videos

  • MOTION IN A PLANE

    NCERT FINGERTIPS|Exercise Relative Velocity In Two Dimensions|4 Videos

  • MOTION IN A PLANE

    NCERT FINGERTIPS|Exercise Vector Addition - Analytical Method|11 Videos

  • MECHANICAL PROPERTIES OF SOLIDS

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

  • MOTION IN A STRAIGHT LINE

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

A particle starts moving from the point (1, 2, 3) under velocity 1 hat I +2hat j + 3hatk for two seconds. After that it moves under velocity 1 hat I -2hat j + 3hatk for the next two seconds. What is the coordinate of the final points?

A particle moves in the x-y the accoding to the equatio, r = (hati + 2hatj) A cos omegat . The motion of the particle is

A particle is acted upon by constant forces 4hati +hatj - 3hatk and 3hati + hatj -hatk which displace it from a point hati + 2hatj + 3hatk to the point 5hati + 4hatj + hatk . The work done in standard units by the forces is given by:

A uniform force of (3hati+hatj) N acts on a particle of mass 2kg. Hence, the particle is displaced from position (2hati+hatk) m to position (4hati+3hatj-hatk) m. The work done by the force on the particle 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 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 ?