wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The motion of a particle along a straight line is described by the equation x=6+27t−t3.
Find the time when K.E of the particle is zero.

A
3 s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 3 s
The motion of the particle is given by x=6+27tt3
Its velocity, v=273t2
Kinetic energy becomes zero when the velocity becomes zero.
v=273t2=0
t=+3,3
Time is always positive.
The K.E of the particle becomes zero at t=3 s

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