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

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B

9

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C

10

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D

1.8

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Solution

## The correct option is B 9Step (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$v–u=at\phantom{\rule{0ex}{0ex}}a=\frac{\left(v–u\right)}{t}\phantom{\rule{0ex}{0ex}}a=\frac{\left(20–10\right)}{t}\phantom{\rule{0ex}{0ex}}a=\frac{10}{t}$Step (3) Finding the time taken by the particle:From the equations of motion, the distance covered by the particle is given by $s=ut+\left(\frac{1}{2}\right)a{t}^{2}\phantom{\rule{0ex}{0ex}}Bysubstitutingthecorrespondingvalues,weget\phantom{\rule{0ex}{0ex}}135=10t+\left(\frac{1}{2}\right)\left(\frac{10}{t}\right){t}^{2}\phantom{\rule{0ex}{0ex}}135=10t+5t\phantom{\rule{0ex}{0ex}}Weget,\phantom{\rule{0ex}{0ex}}t=9s$Hence, the answer is (B) 9 s

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