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

A bus moves from stop A to the next stop B. Its acceleration varies as a=αβx, where α and β are positive constants and x is the distance from the stop A to stop B. The distance between A and B and the maximum speed of the bus are

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

The correct option is A 2αβ and αβ
The acceleration equation a=αβx can be written as v dv=(αβx)dx The bus starts from rest at the stop A. By integrating, we get
v0v dv=x0(αβx)dx
v2=2αxβx2
Substitute v=0 (the bus comes to rest at the stop B) to get the distance between two stops.
Distance between two stops = x=2αβ


The velocity is maximum when acceleration (first derivative of velocity) is zero. Substitute a=0 in a=αβx to get x=αβ, the position of maximum velocity.
Substitute x=αβ in v2=2αxβx2 to get the maximum velocity v=αβ

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