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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
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B
αβ and αβ
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C
αβ and 2αβ
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D
2αβ and 2αβ
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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

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