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The relation between time t and distance x for a moving body is given as t = mx² + nx, here m and n are constants. The retardation of the motion is: (When v stands for velocity)

1. 2 mv³
2. 2mnv³
3. 2 nv³
4. 2n²v³

Correct answer: 2 mv³

Solution

From t=mx^2+nx, dt/dx=2mx+n=1/v. Then a=v dv/dx; dv/dx=-2m v^2, so a=-2m v^3. The retardation is 2m v^3.

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