Home
Class 12
PHYSICS
A particle is moving such that its posit...

A particle is moving such that its position coordinates `(x, y)` are `(2m, 3m)` at time `t=0, (6m, 7m)` at time `t=2 s`, and `(13 m, 14m)` at time `t=5 s`.
Average velocity vector`(vec(V)_(av))` from `t=0` to `t=5 s` is

A

`(1)/(5)(13 overset ^ i+ 14 overset ^j)`

B

`(7)/(3)(overset ^ i + overset ^j)`

C

`2( overset ^i + overset ^j)`

D

`(11)/(5)(overset ^ i + overset ^ j)`

Text Solution

Verified by Experts

The correct Answer is:
A

`overset rarr v_(w//n)=(Delta overset rarr r)/(Delta t)=((13-2)overset ^i+(14-3)overset^j)/(5-0)=(11)/(5)(overset ^i + overset ^j)`
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    PHYSICS WALLAH|Exercise LEVEL-2|30 Videos
  • MOTION IN A PLANE

    PHYSICS WALLAH|Exercise NEET Past 5 years Questions|10 Videos
  • MECHANICAL PROPERTIES OF MATTER

    PHYSICS WALLAH|Exercise NEET Past 5 Yeats Questions|16 Videos
  • MOTION IN A STRAIGHT LINE

    PHYSICS WALLAH|Exercise NEET Past 5 Years Questions|5 Videos

Similar Questions

Explore conceptually related problems

A particle is moving such that its position coordinates (x,y) are (2m,3m) at time t=0,(13m,14m) at time t=5s .Average velocity vector from t =0 to t=5s

A particle is moving along x-axis such that its velocity varies with time according to v=(3m//s^(2))t-(2m//s^(3))t^(2) . Find the velocity at t = 1 s and average velocity of the particle for the interval t = 0 to t = 5 s.

A particle is moving with uniform acceleration. Its position (x) is given in term of time (t in sec) as x= '(5t^2 +4t +8)' m, then

A particle is moving on a straight line. Its velocity at time t is (8-2t)m//s . What is the total distance covered from t=0 to t=6s ?

A particle moving along the x axis has a position given by x=54 t - 2.0 t^(3) m . At the time t=3.0s , the speed of the particle is zero. Which statement is correct ?

A particle moves along X-axis in such a way that its coordinate X varies with time t according to the equation x = (2-5t +6t^(2))m . The initial velocity of the particle is

Two particles of masses m_(1) and m_(2) in projectile motion have velocities vec(v)_(1) and vec(v)_(2) , respectively , at time t = 0 . They collide at time t_(0) . Their velocities become vec(v')_(1) and vec(v')_(2) at time 2 t_(0) while still moving in air. The value of |(m_(1) vec(v')_(1) + m_(2) vec(v')_(2)) - (m_(1) vec(v)_(1) + m_(2) vec(v)_(2))|

A particle is moving with velocity v=4t^(3)+3 t^(2)-1 m//s . The displacement of particle in time t=1 s to t=2 s will be

Find displacement of a particle in 1-D if its velocity is v=(2t-5) m//s , from t=0 to t=4 sec.