To solve this, we can use vector addition and trigonometry to find the total displacement of the plane. The key idea is to break the journey into two parts, then add the vector components of each part to find the total displacement.

Step-by-step breakdown:

1. Convert flight times into distances:

   • First leg: $1.3 \, \text{hours} \times 110 \, \text{mph} = 143 \, \text{miles}$
   • Second leg: $1.5 \, \text{hours} \times 110 \, \text{mph} = 165 \, \text{miles}$

2. Break each leg into components:

   • For the first leg, we use the bearing $40^\circ$.

     • Northward component: $143 \times \cos(40^\circ)$
     • Eastward component: $143 \times \sin(40^\circ)$

   • For the second leg, we use the bearing $130^\circ$.

     • Northward component: $165 \times \cos(130^\circ)$
     • Eastward component: $165 \times \sin(130^\circ)$

3. Sum the components:

   • Total northward displacement: Sum of north components from both legs.
   • Total eastward displacement: Sum of east components from both legs.

4. Calculate the total distance from the starting point using the Pythagorean theorem: $\text{Total distance} = \sqrt{(\text{total northward displacement})^2 + (\text{total eastward displacement})^2}$

I'll calculate the result now.The plane is approximately 218.34 miles from its starting point.

Would you like further details or have any questions about this solution?

Here are five related questions for practice:

1. If the plane flew at 120 mph instead, how far would it be from its starting point?
2. What is the bearing of the plane’s final position relative to its starting point?
3. How would the result change if the plane flew on the second leg at a bearing of $150^\circ$?
4. What if the first leg was flown for 1.5 hours and the second leg for 1.3 hours at the same speed?
5. How far would the plane travel if it flew straight back to its starting point after the second leg?

Tip: Bearings are always measured clockwise from the north.