Question

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

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

Answer 1

The correct option is D $-\frac{20}{3}m{s}^{-2}$

In this figure,

The relation between $v$ & $x$ comes to be

$V=\frac{-20}{30}\times x+20$

now differentiate with respect to time $\left(t\right)$

we get $\frac{dv}{dt}=\left(\frac{-20}{30}\right)\times \frac{dx}{dt}$

as $\frac{dv}{dt}=a\&\frac{dx}{dt}=v$

put the respective values in the equation

$\begin{array}{c}=>a=\left(\frac{-20}{30}\right)v\\ =>a=\left(\frac{-20}{30}\right)\times \left(\frac{-20}{30}\times x+20\right)\end{array}$

now to find a at $x=15$

put $x=15$ in above equation

we get $a=\frac{-20}{3}$

$\therefore$ Option $D$ is correct.