Answer
It takes a point on the front of the train 28.6 seconds to go around this portion of the track.
Work Step by Step
We can convert the train's speed to feet per second:
$v = 30.0~mph\times \frac{1~hr}{3600~s}\times \frac{5280~ft}{1~mi} = 44~ft/s$
We can convert the central angle to radians:
$\theta = (40^{\circ})(\frac{\pi~rad}{180^{\circ}}) = 0.698~rad$
We can find the distance $d$ of the arc length:
$d = \theta ~r$
$d = (0.698~rad)(1800~ft)$
$d = 1256.4~ft$
We can find the time it takes the train to complete this distance:
$t = \frac{d}{v}$
$t = \frac{1256.4~ft}{44~ft/s}$
$t = 28.6~s$
It takes a point on the front of the train 28.6 seconds to go around this portion of the track.
