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

A body starts from rest with acceleration 2 m/s2 till it attains the maximum velocity, then retards to rest with 3 m/s2. If total time taken is 10 seconds then maximum speed attained is

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

The correct option is A 12 m/s
I - Method
Let the maximum velocity attained at time (t) be vmax.
By first equation of motion

vmax=u+at vmax=2t ...(i)

Now for the second part of the journey body comes to rest (v=0) and the time will be (10t) as the total journey time is 10 s, u=vmax, a=3 m/s2. By first equation of motion

0=vmax3(10t) ...(ii) putting the value of vmax in equation (ii) we get,
2t=3(10t), t=6 s
vmax=2t=2×6=12 m/s

IIMethod
As we know that slope (tanθ) of velocity-time (vt) graph gives acceleration.
tanα=2, tanβ=3
2=vt, 3=v10t
v=2t, v=303t
303t=2tt=6 sec
v=12 m/s

flag
Suggest Corrections
thumbs-up
59
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