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

The position of the particle is given by x=3t26t+4 in m. The acceleration of the particle at 10 sec is

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

The correct option is B 6 m/s2
As we know that the acceleration of the particle is given by
a=dvdt and also, v=dxdt
Given, x=3t26t+4
So, v=dxdt=d(3t26t+4)dt
v=6t6
also, a=dvdt=d(6t6)dt=6
Thus, the acceleration of the particle is constant i.e 6 m/s2

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