#### Answer

Yes, see explanations.

#### Work Step by Step

Step 1. With the given conditions, we have
$b=15\ ft, 2a=50, a=25\ ft$,
Step 2. The center is at $(0,0)$, we can write the equation as
$\frac{x^2}{625}+\frac{y^2}{225}=1$
Step 3. The width of the truck gives $x=14\ ft$; plug-in the equation and we have $\frac{14^2}{625}+\frac{y^2}{225}=1$, which gives $y\approx12.4\ ft$
Step 4. The height of the truck is $12 \lt 12.4\ ft $; this means that the truck can pass under the archway without going into the other lane.