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 `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, 1^c = `(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.)