Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

The position vector of a particle R→ as a function of time is given by R→=4sin(2πt)i^+4cos(2πt)j^Where R is in meters, t is in seconds and i^ and j^ denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

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

b

Path of the particle is a circle of radius } 4 \text { meter.

c

Acceleration vector is along -R→ .

d

Magnitude of acceleration vector is v2R, where v is the velocity of particle.

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Here, R→=4sin(2πt)i^+4cos(2πt)j^The velocity of the particle isv→=dR→dt=ddt[4sin(2πt)i^+4cos(2πt)j^]=8πcos(2πt)i^-8πsin(2πt)j^its magnitude is|v→|=(8πcos(2πt))2+(-8πsin(2πt))2=64π2cos2(2πt)+64π2sin2(2πt)=64π2cos2(2πt)+sin2(2πt)=64π2                                        as sin2θ+cos2θ=1=8πm/s
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring