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The relation between time t and distance...

The relation between time `t` and distance `x` is `t = ax^(2)+ bx` where `a and `b` are constants. The acceleration is

A

`2bv^(3)`

B

`2av^(3)`

C

`-2av^(2)`

D

`-2av^(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`t = ax^(2) + bx`
Differentiate the equation with respect to t
`therefore 1 = 2ax (dx)/(dt) + b (dx)/(dt)`
`v = 1/(2ax + b)`
or `(dv)/(dt) = (-2a (dx//dt))/(2ax + b)^(2) = -2av xx v^(2)`
Acceleration `=-2av^(3)`
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