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The relation between time $t$ and distance $\mathrm{x}$ is $\mathrm{t}=a \mathrm{x}^2+\mathrm{bx}$ where $\mathrm{a}$ and $\mathrm{b}$ are constants. The acceleration is
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Verified Answer
The correct answer is:
$-2 a v^3$
$-2 a v^3$
$$
t=a x^2+b x
$$
by differentiating acceleration $=-2 a v^3$
t=a x^2+b x
$$
by differentiating acceleration $=-2 a v^3$
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