CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Starting from rest a particle moves in a straight line with acceleration
a=(25t2)1/2m/s2 for 0t5s
a=3π8m/s2 for t5s
The velocity of particle at t = 7s is:

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

The correct option is B 22 m/s
v=adt

a=(25t2)1/2m/s2 for 0t5s
=3π8m/s2 for t5s

v=(25t2)1/2dt+a 0t5s
=3π8dt+b t5s

v=t252t2+252sin1(t5)+a 0t5s
=3πt8+b t5s

at t=0,v=0 gives a=0
finding b using the fact that speed is a continous fuction
limt5v=limt5+v
0+252sin1(55)=3π×58+b
b=35π8

v=t252t2+252sin1(t5) 0t5s
=3πt8+35π8 t5s

so at t=7
v=3π×78+35π8
v=8π×78
v=7π
v=22m/s





flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon