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

Options

A. `2bv^(3)`
B. `2av^(3)`
C. `-2av^(2)`
D. `-2av^(3)`

Answer

D

Explanation

`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)`