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

Options

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

Answer

3

Explanation

Given `t=ax^(2)+bx`
Differentiating w.r.t 't'
`(dt)/(dt)=2ax (dx)/(dt)+b(dx)/(dt)`
`upsilon=(dx)/(dt)=1/((2ax+b))`
Again differentiating w.r.t 't'
`(d^(2)x)/(dt^(2)) =(d(2ax+b)^(-1))/(d(2ax+b)).2a(dx)/(dt)`
`:. F=(d^(2)x)/(dt^(2))=1/((2ax+b)^(2)) . (2a)/((2ax+b))`
or `f=(-2a)/((2ax+b)^(3))`
`:. f=-2av^(3)`