If the velocity v of a particle moving along a straight line decreases linearly with its displacement s from $20 m s^{- 1}$ to a value approaching zero at s = 30 m, then acceleration of the particle at s = 15 m is

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$\frac{2}{3} m s^{- 2}$

$- \frac{2}{3} m s^{- 2}$

$\frac{20}{3} m s^{- 2}$

$- \frac{20}{3} m s^{- 2}$

answer is D.

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Detailed Solution

Slope of line $= - \frac{2}{3}$

Equation of line is $(υ - 20) = - \frac{2}{3} (s - 0)$

$\Rightarrow υ = 20 - \frac{2}{3} s ................. (i)$

Velocity at s = 15 m, i.e.,

$v = \frac{d s}{d t} |_{s = 15 m} = 20 - \frac{2}{3} (15) = 10 m s^{- 1}$

$∴ a = \frac{d v}{d t} |_{s = 15 m} = - \frac{2}{3} \frac{d s}{d t} |_{s = 15 m}$

$= - \frac{20}{3} m s^{- 2}$

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