Question

A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 \, m \, s^{- 1}$ to $20 \, m \, s^{- 1}$ while passing through a distance $135 \, m$ in $t$ seconds. The value of $t$ is:

• A. $10$
• B. $1.8$
• C. $12$
• D. $9$

Answer 1

Answer: (D)

Initial velocity, $u \, = \, 10 \, m \, s^{- 1}$
Final velocity , $v \, = \, 20 \, m \, s^{- 1}$
Distance , $s \, = \, 135 \, m$
Let , acceleration = a
Using the formula , $v^{2} \, = \, u^{2} + 2 a s$
$\Rightarrow \, \, \, a \, = \, \frac{v^{2} - u^{2}}{2 s}$
$or \, \, a = \, \frac{\left(\right. 20 \left.\right)^{2} - \left(\right. 10 \left.\right)^{2}}{2 \times 135} \, = \, \frac{400 - 100}{2 \times 135}$
$or \, \, a \, = \, \frac{300}{2 \times 135} \, = \, \frac{150}{135} \, = \, \frac{10}{9} \, m \, s^{- 2}$
Now , using the relation , $v \, = \, u + a t$
$t=\frac{v-u}{a}=\frac{20-10}{\frac{10}{9}}=\frac{10}{10}\times 9=9\mathrm{s}.$