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

A particle moves along xaxis in positive direction. Its acceleration a is given as a=cx+d, where x denotes the xcoordinate of particle, c and d are positive constants. For velocity-position graph of particle to be of type as shown in figure, the value of speed of particle at x=0 should be:
1021948_11ef4f87d816421f9d6d119089ab2c6f.png

A
4d2c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
d2c
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2d2c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8d2c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B d2c
a=cx+d
u=kx+u..(1)
u.dvdx=cx+d
vk=cx+d
v=kcxk2+dk2...(2)
Compare equation (1) and (2)
k=c
u=dk=dc

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Motion Under Constant Acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon