If the velocity $v$ of a particle moving along a straight line decreases linearly with its displacement $s$ from 20$ms^{- 1}$ to a value approaching zero $s = 30$ m, then acceleration of the particle at $s = 15$m is [image:3_image119.emf]

Question

Dr. Abubacker · Accepted Answer

(d) Slope of line $= - \frac{2}{3}$
Equation of line is $(v - 20) = - \frac{2}{3}(s - 0)$
$\Rightarrow \ v = 20 - \frac{2}{3}s$ …(i)
Velocity at $s = 15m\ ie,$
$v = \left. \ \frac{ds}{dt} ight|_{s = 15m}$ $= 20 - \frac{2}{3}(15) = 10ms^{- 1}$
Differentiate Eq. (i) with respect to time,
acceleration $= \frac{dv}{dt} = \frac{2}{3}\frac{ds}{dt}$
$	herefore\ \left. \ \frac{dv}{dt} ight|_{s = 15m} = - \frac{2}{3}\ \left. \ \frac{ds}{dt} ight|_{s = 15m} = - \frac{20}{3}ms^{- 2}$

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

Explanation

Reviewed by Dr. Abubacker

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