Question

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms^-1 to 20 ms^-1 while passing through a distance of 135 m in ‘t’ seconds. The value of t is:

• A. 12
• B. 9
• C. 10
• D. 1.8

Answer 1

The correct option is B

9

Step (1) Given:

The initial velocity, $u=10m/s$

The final velocity, $v=20m/s$

Distance covered, $s=130m$

Step (2) Finding the acceleration of the particle:

From the equations of motion, we know that

$$\begin{gathered} v–u=at \\ a=\frac{(v–u)}{t} \\ a=\frac{(20–10)}{t} \\ a=\frac{10}{t} \end{gathered}$$

Step (3) Finding the time taken by the particle:

From the equations of motion, the distance covered by the particle is given by

$$\begin{gathered} s=ut+\left(\frac{1}{2}\right)a{t}^2 \\ Bysubstitutingthecorrespondingvalues,weget \\ 135=10t+\left(\frac{1}{2}\right)\left(\frac{10}{t}\right)t^2 \\ 135=10t+5t \\ Weget, \\ t=9s \end{gathered}$$

Hence, the answer is (B) 9 s