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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.
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Solution
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 `1/2` minute
= `36/60 xx 1/2`
= `3/10 "km"`
Let θ be the required angle.
Then S = rθ
where S = distance travelled in 30 seconds
= `3/10"km"` and r = 1 km
∴ `3/10` = 1 x θ
∴ θ = `(3/10)^"c"`
Now, 1c = `(180/pi)^circ`
∴ `(3/10)^"c" = (3/10 xx 180/pi)^circ`
= `(54/pi)^circ`
= `((54 xx 7)/22)^circ`
= 17.18°
= 17° + 0.18°
= 17° + (0.18 x 60)'
= 17° + (10.8)'
= 17°11' ...(approx.)
Hence, the required angle = 17°11' ...(approx.)
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