Magnitude of Average Velocity

Question

A train travels 100 miles toward 37 degrees northwest and then 90 miles north. The whole journey takes 2 hours. What is the magnitude of the average velocity of the train?

Answer 1

$\displaystyle\frac{\sqrt{(100\cos 37)^2+(100\sin 37+90)^2}}{2}\approx85.05$

1. The cosinus is the x/east component of the journey.
2. The sinus is the first part of the y/north component and is added to the second part of the jouney, that went straight to north.
3. The square root is Pythagoras theorem on the components of the distances $d=\sqrt{ x^2+y^2}$.
4. The divisor $2$ is the time and $\displaystyle v=\frac{d}{t}$.

Answer 2

I think I've figured it out after having a rest!

I constructed the vector diagram, discovered that I have two sides and an angle. From there I let delta(x) = sqrt[100^2 + 90^2 - 2(100)(90)cos(127)] = 170 mi. Average velocity = delta(x)/delta(t) = 170 mi/2 hr = 85 mi/hr.