(a) This first question is asking us to construct an equation for the distance travelled by the car and the distance travelled by the plane. Let's say that both the car and the plane traveled for 2 hours:

Distance traveled by car = 60km/h*2hours = 120km

Distance traveled by plane = 650km/h*2hours = 1300km

So if we say that both the car and plane travelled for t hours:

**r = 60t**

**s = 650t**

(b) To do this, we will be pulling out our trigonometry, specifically the pythagorean formula. For now, lets assume that the plane and car are both at ground level. Assuming both the car and the plane start at the same location, a right triangle can be created from the final distances the plane and car go after some time. The three points of this triangle are where both the plane and car started, where the plane ended, and where the car ended. The formula for the distance between these two points is the hypotenuse of this triangle, which can be found in the following way:

d_{1}^{2} = r^{2} + s^{2}

Now, this distance assumes that the plane is at ground level, which it isn't. To account for the fact the plane is 4km above ground level, we can solve for the hypotenuse of another right triangle. This time, the three points that make up the triangle are the end location of the car, the end location of the plane assuming it is at ground level, and the actual end location of the plane. This would make the following equation:

d^{2} = d_{1}^{2} + 4^{2}

Let's substitute d_{1} and make the equation equal to d.

**d = sqrt(r**^{2}** + s**^{2}** + 16)**

(c) Let's substitute our equations from (a) into our equation from (b)

d = sqrt((60t)^{2} + (650t)^{2} + 16)

**d = sqrt(426100t**^{2}**+16)**

(d) If they start at noon and end at 1:30PM, then the plane and car travelled for 1.5 hours. Let's plug this in for t in our equation formed in (c).

d = sqrt(426100(1.5)^{2}+16)

**d = 1,130.6km**

Jirecho C.

Thank you so much. God Bless10d