Question

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

• A. $\frac{2}{3} \text{ m s}^{- 2}$
• B. $- \frac{2}{3} \text{ m s}^{- 2}$
• C. $\frac{20}{3} \text{ m s}^{- 2}$
• D. $- \frac{20}{3} \text{ m s}^{- 2}$

Answer 1

Answer: (D)

Slope of line $= - \frac{2}{3}$
Equation of line is $\left(\text{v} - 2 0\right) = - \frac{2}{3} \left(\text{S} - 0\right)$
$⇒ \, \, \text{V} = 2 0 - \frac{2}{3} \text{S} \, \, … \left(\text{i}\right)$
Velocity at S = 15 m,
i.e., $v=20-\frac{2}{3}\left(15\right)=10 \, m \, s^{- 1}$
$a=v\frac{d v}{d s}=v\times slope$

$a=10\left(- \frac{2}{3}\right)$

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