Home
Class 12
PHYSICS
The position vector of a particle vecR a...

The position vector of a particle `vecR` as a function of time is given by:
`vecR=4sin(2pit)hati+4cos(2pii)hatj`. Where R is in meters,t is in seconds and `hati` and `hatj` denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?

A

Path of the particle is a circle of radius 4 meter

B

Acceleration vectors is along `-vecR`

C

Magnitude of acceleration vector is `(v^(2))/(R)` where v is the velocity of particle

D

Magnitude of the velocity of particle is 8 meter/ second.

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (Oblique projection )|25 Videos

  • MAGNETISM

    AAKASH SERIES|Exercise PROBLEMS (LEVEL-II)|13 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise QUESTION FOR DESCRIPTIVE ANSWER|7 Videos

Similar Questions

Explore conceptually related problems

The position vector of a particle is vecr = (a cos omega t)hati + (a sin omegat)hatj . the velocity of the particle is

The position vector of a particle moving in a plane is given by r =a cos omega t hati + b sin omega I hatj, where hati and hatj are the unit vectors along the rectangular axes C and Y : a,b, and omega are constants and t is time. The acceleration of the particle is directed along the vector

The position vector of a moving particle at't' sec is given by vecr = 3hati + 4t^(2)hatj - t^(3)hatk . Its displacement during an interval of t = Is to 3 sec is

The position vector of a particle is given by r = 3t^(2)hati + 4t^(2)hatj + 7hatk m at a given time 't'. The net displacement of the particle after 10s is

The position of a particle axecuting simple harmonic motion is given by x (t) = 2 cos ((pi)/(15) t - (pi)/(2)), where x is in centimetre and t is in seconds. The time period of the kinetic energy of the particle in second is