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

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Aicharya · Accepted Answer

Key Idea : The problem requires kinematics equations of motion.
Let $u$ and $v$ be the first and final velocities of particle and $a \& s$ be the constant acceleration and distance covered by it.
From third equation of motion
$\begin{array}{l}
y^2 =u^2+2 a s \\
\Rightarrow (20)^2 =(10)^2+2 a \times 135 \\
\text {or } a =\frac{300}{2 \times 135}=\frac{10}{9} \mathrm{~ms}^{-2}
\end{array}$
Now using first equation of motion,
$\begin{aligned}
& v=u+a t \\
& \text {or } t=\frac{v-u}{a}=\frac{20-10}{(10 / 9)}=\frac{10 \times 9}{10}=9 \mathrm{~s}
\end{aligned}$

A particle moves in a straight line with a constant acceleration. It changes its velocity from $10…

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