Question

If the velocity $v$ of a particle moving along a straight line decreases linearly with its displacement $S$ is $20m{s}^{-1}$ to a value approaching to zero at $S=30m$ , then the acceleration of the particle at $S=15m$ , 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 $\left(\text{v}-20\right)=-\frac{2}{3}\left(\text{S}-0\right)$
$\Rightarrow \text{V}=20-\frac{2}{3}\text{S}\dots \left(\text{i}\right)$
Velocity at S = 15 m,
i.e., $v=20-\frac{2}{3}\left(15\right)=10m{s}^{-1}$
$a=v\frac{dv}{ds}=v\times slope$

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

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