MCQ
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..........$s$
  • A
    $12 $
  • $9$
  • C
    $10$
  • D
    $1.8$

Answer

Correct option: B.
$9$
b
$\begin{array}{l}
{v^2} - {u^2} = 2\,as\\
Give\,v = 20\,m{s^{ - 1}},\,u = 10\,m{s^{ - 1}},\,s = 135\,m\\
\therefore a = \frac{{400 - 100}}{{2 \times 135}} = \frac{{300}}{{270}} = \frac{{10}}{9}m/{s^2}\\
\,\,\,\,\,v = u + at\, \Rightarrow \,t = \frac{{v - u}}{a} = \frac{{10m/s}}{{\frac{{10}}{9}m/{s^2}}} = 9s
\end{array}$

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