Motion in a Straight Line - Result Question 20
20. 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 $135 m$ in $t$ second. The value of $t$ is:
[2008]
(a) 10
(b) 1.8
(c) 12
(d) 9
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Answer:
Correct Answer: 20. (d)
Solution:
- (d) Initial velocity, $u=10 ms^{-1}$
Final velocity, $v=20 ms^{-1}$
Distance, $s=135 m$
Let, acceleration $=a$
Using the formula, $v^{2}=u^{2}+2 a s$
$a=\frac{v^{2}-u^{2}}{2 s}$
$=\frac{(20)^{2}-(10)^{2}}{2 \times 135}=\frac{400-100}{2 \times 135}$
or, $a=\frac{300}{2 \times 135}=\frac{150}{135}=\frac{10}{9} ms^{-2}$
Now, using the relation, $v=u+a t$
$t=\frac{v-u}{a}=\frac{20-10}{10 / 9}=\frac{10}{10} \times 9 sec=9 sec$.
