#### 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.