A point moves rectilinearly with deceleration which depends on the velocity v of the particle as a=kv where k=+ve constant. At the initial moment the velocity of the point is equal to v0. What distance will it cover before its stops.

a=KV;∫V12dv=∫0skds;s=23V03/2K

A point moves rectilinearly with deceleration which depends on the velocity v of the particle as a=kv where k=+ve constant. At the initial moment the velocity of the point is equal to v0. What distance will it cover before its stops.

a=KV;∫V12dv=∫0skds;s=23V03/2K

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A point moves rectilinearly with deceleration which depends on the velocity v of the particle as $a = k \sqrt{v}$ where k=+ve constant. At the initial moment the velocity of the point is equal to $v_{0} .$ What distance will it cover before its stops.

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$s = \frac{2 v_{0}^{3 / 2}}{3 k}$

$s = \frac{2 v_{0}^{1 / 2}}{3 k}$

$s = \frac{2 v_{0}^{}}{7 k}$

$s = \frac{2 v_{0}^{3 / 2}}{k}$

answer is A.

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$a = K \sqrt{V}; \int V^{\frac{1}{2}} d v = \int_{0}^{s} k d s; s = \frac{2}{3} \frac{V_{0}^{3 / 2}}{K}$

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A point moves rectilinearly with deceleration which depends on the velocity v of the particle as a=kv where k=+ve constant. At the initial moment the velocity of the point is equal to v0. What distance will it cover before its stops.

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A point moves rectilinearly with deceleration which depends on the velocity v of the particle as $a = k \sqrt{v}$ where k=+ve constant. At the initial moment the velocity of the point is equal to $v_{0} .$ What distance will it cover before its stops.