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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

`2A (Ax + B)^(–3)`

B

`2A (2Ax + B)^(-3)`

C

`A//(Ax + B)^(-3)`

D

`A//2 [2Ax + B]^(-3)`

Text Solution

Verified by Experts

As ` t = Ax^(2) + Bx rArr dt //dx = 2Ax + B `
` rArr v = (2Ax + B)^(-1)` …(A)
[as dx/dt = v], Now by chain rule
` a= (dv)/(dt) = (dv)/(dx) .(dx)/(dt) = v(dv)/(dx)`
` rArr a = (2Ax + B) ^(-1) (d)/(dx) (2Ax + B)^(-1)`
`= - 2 A ( 2Ax + B)^(-3)`
So retardation ` = - a = 2A (2Ax + B)^(-3)`
Hence correct answer is (C).
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