Question

Answer the following :

A train is running on a circular track of radius 1 km at the rate of 36 km per hour. Find the angle, to the nearest minute, through which it will turn in 30 seconds.

Answer 1

Distance travelled by the train on the circular track in 1 hour, i.e., in 60 minutes = 36 km

∴ distance travelled in 30 seconds, i.e., in ${\displaystyle \frac{1}{2}}$ minute

= ${\displaystyle \frac{36}{60}\times \frac{1}{2}}$

= ${\displaystyle \frac{3}{10}\text{km}}$

Let θ be the required angle.

Then S = rθ

where S = distance travelled in 30 seconds

= ${\displaystyle \frac{3}{10}\text{km}}$ and r = 1 km

∴ ${\displaystyle \frac{3}{10}}$ = 1 x θ

∴ θ = ${\displaystyle {\left(\frac{3}{10}\right)}^{\text{c}}}$

Now, 1^c = ${\displaystyle {\left(\frac{180}{\pi }\right)}^{\circ }}$

∴ ${\displaystyle {\left(\frac{3}{10}\right)}^{\text{c}}={(\frac{3}{10}\times \frac{180}{\pi })}^{\circ }}$

= ${\displaystyle {\left(\frac{54}{\pi }\right)}^{\circ }}$

= ${\displaystyle {\left(\frac{54\times 7}{22}\right)}^{\circ }}$

= 17.18°

= 17° + 0.18°

= 17° + (0.18 x 60)'

= 17° + (10.8)'

= 17°11' ...(approx.)

Hence, the required angle = 17°11' ...(approx.)