Question

A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 ms^{-1}$ to $20ms^{-1}$ while passing through a distance 135 m in t second. The value of t is

• A. 10
• B. 1.8
• C. 12
• D. 9

Answer 1

Answer: (D)

9

Answer 2

Let $u$ and $v$ be the first and final velocities of particle and $a$ and $s$ be the constant acceleration and distance covered by it. From third equation of motion
$v^2=u^2+2as$
$\Rightarrow$ $(20)^2=(10)^2+2a \times 135$
or $a=\frac{300}{2 \times 135}=\frac{10}{9}ms^{-2}$
Now using first equation of motion,
v=u+at
or $t=\frac{v-u}{a}=\frac{20-10}{(10/9)}=\frac{10 \times 9}{10}=9s$