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The relation between time t and distance x is given by, t = ax^2 + bx where a and b are constants.
Let's differentiate t with respect to x,
i.e., dt/dx = d(ax^2 + bx) /dx
Or, dt/dx = 2ax + b....(1)
We know, Velocity is the rate of change of displacement with respect to time.
i.e.,v = dt/dx
From equation (1),
dt/dx = 1/V = 2ax + b
Or, V = 1/(2ax +b) c....(2)
Now differentiating V with respect to time t,
dv/dt = d{1/(2ax + b)} /dt
= - 1/(2ax +b) ^2 × d(2ax + b) /dt
= - 1/(2ax +b) ^2 × [2a × dx/dt]
= - 1/(2ax +b) ^2 × 2ab
From equation (2),
dv/dt = - v^2 × 2av = - 2av^3
We know, Accerleration /retardation is the rate of change of Velocity with
respect of time.
i.e., A = dv/dt
So, dv/dt = A = - 2av^3
[Here negative sign shows retardation.]
Hence, answer is - 2av^3.
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