First slide

The 3 equation for uniform acc

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 135 m in t second. The value of t (in second) is

Moderate

A

12
B

9
C

10
D

1.8

Solution

From third equation of motion, $v^{2} = u^{2} + 2 a s$
$\Rightarrow a = \frac{v^{2} - u^{2}}{2 s} = \frac{(20)^{2} - (10)^{2}}{2 \times 135} = \frac{300}{270} = \frac{10}{9} {ms}^{- 2}$
From first equation of motion,
$v = u + a t \Rightarrow t = \frac{v - u}{a} = \frac{20 - 10}{10 / 9} = \frac{10}{10 / 9} = 9 s$

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