Question

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-' in 10 minutes. Find its acceleration​

Answer 1

$\huge {\mathtt{Given:-}}$

$\sf {9}^{x} \times {3}^{2} \times ( {3}^{ \frac{x}{ - 2} } )^{ - 2} = \frac{1}{27}$

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$\huge {\mathtt{To \: Find:-}}$

$\sf value \: of \: x$

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$\huge {\mathtt{Let's \: Do:-}}$

$\sf {9}^{x} \times {3}^{2} \times ( {3}^{ \frac{x}{ - 2} } )^{ - 2} = \frac{1}{27}$

$\sf {(3}^{2} )^{x} \times {3}^{2} \times ( {3}^{ \frac{x}{ - 2}}) ^{ - 2} = \frac{1}{ {3}^{3} }$

$\sf {3}^{2 × x} \times {3}^{2} \times {3}^{ \frac{x}{\cancel{ - 2}} × \cancel{- 2}} = {3}^{ - 3}$

$\sf {3}^{2x} \times {3}^{2} \times {3}^{x} = {3}^{ - 3}$

$\sf {3}^{ 2x } \times {3}^{2} \times {3}^{x} = {3}^{ - 3}$

$\sf {3}^{2x + 2 + x} = {3}^{ - 3}$

$\sf {3}^{3x + 2} = {3}^{ - 3}$

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Now, the bases are same, so the powers will be equal :-

$\sf \implies 3x + 2 = - 3$

$\sf \implies 3x = - 3 - 2$

$\sf \implies 3x = - 5$

$\sf \implies x = - \frac{5}{3}$

$\boxed{\sf x = - \frac{5}{3}}$

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$\sf \therefore \: value \: of \: x = - \frac{5}{3}$

Answer 2

Explanation:

Given: t=10 min=10×60=600s

Initial speed of train is given as: u=0 ms−1

Final speed of train is given as: v=40 km h−1=360040×1000=11.1 ms−1

Now acceleration is given by the relation:

a=tv−u=60011.1−0=0.0185 ms−2