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Maths-

General

Easy

Question

If a particle moves according to the law $s equals 6 t to the power of 2 end exponent minus fraction numerator t to the power of 3 end exponent over denominator 2 end fraction$ , then the time at which it is momentarily at rest is

$t = 0$ only
$t = 8$ only
$t = 0, 8$
None of these

The correct answer is: $t equals 0 comma blank 8$

We have,
$s equals 6 t to the power of 2 end exponent minus fraction numerator t to the power of 3 end exponent over denominator 2 end fraction rightwards double arrow fraction numerator d s over denominator d t end fraction equals 12 t minus fraction numerator 3 t to the power of 2 end exponent over denominator 2 end fraction$ and $fraction numerator d to the power of 2 end exponent s over denominator d t to the power of 2 end exponent end fraction equals 12 minus 3 t$

When the particle is momentarily at rest, we have

$fraction numerator d s over denominator d t end fraction equals 0$ and $fraction numerator d to the power of 2 end exponent s over denominator d t to the power of 2 end exponent end fraction not equal to 0$

Now,

$fraction numerator d s over denominator d t end fraction equals 0 rightwards double arrow t equals 0 comma blank t equals 8$

Clearly, $fraction numerator d to the power of 2 end exponent s over denominator d t to the power of 2 end exponent end fraction not equal to 0$ for $t equals 0 comma blank 8$

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