When a 2 kg car driven at 20 m/s on a level road is suddenly put into neutral gear (i.e. allowed to coast), the velocity decreases in the following manner: V=201+(t20)m/s where t is the time in sec. The power (in Watt) required to drive this car at speed 10 m/s on the same road is

P=F→.v→ P=(ma)v a=dvdt v=201+t20 1+t20=20v 120dtdv=20v2 a=v2400 v=10m/s a=−14m/s2 P=2×14×10=5Watt

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When a 2 kg car driven at 20 m/s on a level road is suddenly put into neutral gear (i.e. allowed to coast), the velocity decreases in the following manner:
$V = \frac{20}{1 + (\frac{t}{20})} m / s$ where t is the time in sec. The power (in Watt) required to drive this car at speed 10 m/s on the same road is

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answer is 5.

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Detailed Solution

$P = \vec{F} . \vec{v} P = (m a) v a = \frac{d v}{d t} v = \frac{20}{1 + \frac{t}{20}} 1 + \frac{t}{20} = \frac{20}{v} \frac{1}{20} \frac{d t}{d v} = \frac{20}{v^{2}} a = \frac{v^{2}}{400} v = 10 m / s a = - \frac{1}{4} m / s^{2} P = 2 \times \frac{1}{4} \times 10 = 5 W a t t$