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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 $t$ second. The value of $t$ is -
A
12
B
9
C
10
D
1.8
Detailed Solution
For constant acceleration: $v^2 = u^2 + 2as$
$(20)^2 = (10)^2 + 2 \times a \times 135$
$400 - 100 = 270\,a$
$a = \dfrac{300}{270} = \dfrac{10}{9}$ m s$^{-2}$
Now use $v = u + at$:
$20 = 10 + \dfrac{300}{270}\,t$
$10 = \dfrac{300}{270}\,t$
$t = \dfrac{10 \times 270}{300} = 9$ s
Check with average velocity: $s = \dfrac{u + v}{2}\,t \Rightarrow 135 = 15\,t \Rightarrow t = 9$ s.
$(20)^2 = (10)^2 + 2 \times a \times 135$
$400 - 100 = 270\,a$
$a = \dfrac{300}{270} = \dfrac{10}{9}$ m s$^{-2}$
Now use $v = u + at$:
$20 = 10 + \dfrac{300}{270}\,t$
$10 = \dfrac{300}{270}\,t$
$t = \dfrac{10 \times 270}{300} = 9$ s
Check with average velocity: $s = \dfrac{u + v}{2}\,t \Rightarrow 135 = 15\,t \Rightarrow t = 9$ s.