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A train accelerating uniformly from rest attains a maximum speed of $40 \mathrm{~ms}^{-1}$ in $20 \mathrm{~s}$. It travels at the speed for $20 \mathrm{~s}$ and is brought to rest with uniform retardation in further $40 \mathrm{~s}$. What is the average velocity during the period ?
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The correct answer is:
$25 \mathrm{~m} / \mathrm{s}$
(i) $\mathrm{v}=\mathrm{u}+$ at $_{1}$
$$
\begin{array}{l}
40=0+a \times 20 \\
a=2 m / s^{2} \\
v^{2}-u^{2}=2 a s \\
40^{2}-0=2 \times 2 s_{1} \\
\text { (ii) } s_{1}=400 \mathrm{~m} \\
\text { (iii) } v=u+a t \\
\quad 0=40+a \times 40 \\
a=-1 \mathrm{~m} / \mathrm{s}^{2} \\
0^{2}-40^{2}=2(-1) \mathrm{s}_{3} \\
\mathrm{~s}_{3}=800 \mathrm{~m} \\
\text { Total distance travelled }=\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{3} \\
=400+800+800=2000 \mathrm{~m} \\
\text { Total time taken }=20+20+40=80 \mathrm{~s} \\
\text { Average velocity }=\frac{2000}{80}=25 \mathrm{~m} / \mathrm{s}
\end{array}
$$
$$
\begin{array}{l}
40=0+a \times 20 \\
a=2 m / s^{2} \\
v^{2}-u^{2}=2 a s \\
40^{2}-0=2 \times 2 s_{1} \\
\text { (ii) } s_{1}=400 \mathrm{~m} \\
\text { (iii) } v=u+a t \\
\quad 0=40+a \times 40 \\
a=-1 \mathrm{~m} / \mathrm{s}^{2} \\
0^{2}-40^{2}=2(-1) \mathrm{s}_{3} \\
\mathrm{~s}_{3}=800 \mathrm{~m} \\
\text { Total distance travelled }=\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{3} \\
=400+800+800=2000 \mathrm{~m} \\
\text { Total time taken }=20+20+40=80 \mathrm{~s} \\
\text { Average velocity }=\frac{2000}{80}=25 \mathrm{~m} / \mathrm{s}
\end{array}
$$
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