The speed of a train changes from $36 {km} / {h}$ to $72 {km} / {h}$ in $10 {sec}$. Calculate the distance travelled during this time.

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**ANSWER : 150 meters.**

**SOLUTION ****—**

$v=u+f t ; . \text { or, } v-u=f t $

and $v^2=u^2+2 f s ; \text { or, } v^2-u^2=2 f s$

or, $\frac{v^2-u^2}{v-u}=\frac{2 s}{t} \therefore s=\left(\frac{v+u}{2}\right) t $

Now, $36 {km} / {h}=\frac{36 \times 1000}{60 \times 60}=10 {m} / {sec}$.

and $72 {km} / {h}=\frac{72 \times 1000}{60 \times 60}=20 {~m} / {sec}$.

$s=\left(\frac{v+u}{2}\right) t=\left(\frac{10+20}{2}\right) \times 10=150 \text { metre }$

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