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STD 11 - 2. motion in straight line — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 2. motion in straight line4 Marks
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 $t$ is $..........s$
✓

Answer

$v^2-u^2=2 a s$
Give $v=20\ ms^{-1}, u=10 ms^{-1}, s=135 m$
$\therefore a=\frac{400-100}{2 \times 135}=\frac{300}{270}=\frac{10}{9}\ m / s^2$
$v=u+a t$
$\Rightarrow t=\frac{v-u}{a}$
$=\frac{10 m / s}{\frac{10}{9} m / s^2}$
$=9 s$

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