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Standard XII

Physics

Acceleration

The accelerat...

Question

The acceleration $^{'} a^{'}$ and displacement $^{'} x^{'}$ are related to each other as $a = - x$ . If the velocity of the particle is $u$ at zero displacement., then the distance covered by the particle to come to rest is equal to.

2 u

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Solution

The correct option is D $u$
We know that if $x$ is displacement then we can write the acceleration as $a = v \frac{d v}{d x}$
bit this is given as $a = - x$ so $- x = v \frac{d v}{d x}$
or $- \int_{x = 0}^{X} x d x = \int_{u}^{0} v d v$
integrating above equation we will get $X = u$
so the distacne covered before coming to $r e s t$ is $u .$

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The acceleration ′a′ and displacement ′x′ are related to each other as a=−x. If the velocity of the particle is u at zero displacement., then the distance covered by the particle to come to rest is equal to.

The correct option is D uWe know that if x is displacement then we can write the acceleration as a=vdvdxbit this is given as a=−x so −x=vdvdxor −∫Xx=0xdx=∫0uvdvintegrating above equation we will get X=uso the distacne covered before coming to rest is u.

The acceleration $^{'} a^{'}$ and displacement $^{'} x^{'}$ are related to each other as $a = - x$ . If the velocity of the particle is $u$ at zero displacement., then the distance covered by the particle to come to rest is equal to.