wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

A particle is moving in a circle of radius r=2 m. Its speed with time varies according to the relation v(t)=46t2. What is the magnitude of the total acceleration of the particle at t=2 sec ? Use the value of 769=27.73 for calculation.

A
2.11 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3.46 m/s2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3.1 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.5 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 3.46 m/s2
Given the variation of the speed of the particle is:
v(t)=46t2
Diffrentiating both sides w.r.t time, we get:
dvdt=0+12t3

Now, tangential acceleration at is the rate of change of speed with time,
i.e at=dvdt=12t3
At t=2 s,
at=128=32 m/s2

Now speed at t=2 s:
v(t=2)=464=52 m/s
So, centripetal acceleration at t=2 sec
ac=v2r=(254)2=258 m/s2

Total acceleration aT of particle is
aT=a2c+a2t=62564+94=7698
=27.738=3.46 m/s2


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relative
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon