SEAN and the TYrannosaurus Rex

I used the concept of relative velocity.

The separation between the Sean and Tyrannosaurus decreases at the rate of $2vcos\theta+v$

So $-\frac { dx }{ dt } =(2vcos\theta +v)$

or $\int _{ 0 }^{ T }{ (2vcos\theta +v)dt } =-\int _{ x }^{ x }{ dx }$

Let $T$ be the time when Sean has completed one revolution. Now as they return at their initial position so the distance between will remain unchanged. So the limits of integration of the term in RHS is from $x$ to $x$ .

So $2vcos\theta=-v$ . Now as value of $\theta$ is between $-90$ to $-180$ .

So $\theta=-120$ .

First physics problem, lol, not that good, but what the heck.

Because the T. rex will be constantly traveling at half Sean's speed, the radius of the circle it travels in will be half that of Sean's.

Also, since at all times the direction the Trex travels will be tangent to the circle he travels in, the radius from the center of the circle to the point of tangency (where it begins) will be perpendicular to the t-Rex's initial instantaneous direction. Thus the center of the trex's circle is located at (x,-y).

By observation, the circles they travel in will be concentric (this is very hard for me to explain. Anyone?).

Now, since $\angle S_0T_0C$ (note $C$ denotes both the t-Rex's circle's center and Sean's circle's center) is 90 degrees, we have a right triangle, with hypotenuse $SC$ and leg $T_0C$ . Since $SC$ is twice $T_0C$ , we have a 30-60-90 triangle with the 30 being opposite $T_0C$ . Since tangents are perpendicular to their radius, we have $\theta=90+30=120$

The key to the solution is knowing that SEAN runs twice as fast as the T-Rex. If SEAN and the T-Rex both run around their circles and return back to their starting point at the same time, then the circumference of SEAN's path must be twice as big as the T-Rex's circle. Because $C=2πr$ , the radius of each circle is proportional to it's circumference, meaning that the radius of SEAN's circle is twice that of the T-Rex's. Knowing this, we can add some lines to the picture to make things a little clearer. Let the length of the radius plotted from point C to (x,0) be R. This means that the length of the radius plotted from C to (0,0) is 2r. From here the answer is clear. We now have a right triangle in the 4th quadrant where one of the legs is R (the other being x) and the hypotenuse is 2R. The only kind of triangle that has a hypotenuse that is twice one of it's legs is a 30°-60°-90° triangle. The angle opposite of side R is 30°, which is -30° off of the x-axis, but we're trying to find the angle of the tangent line off of the x-axis. Well, our hypotenuse is a radius, and radii are perpendicular to tangent lines, meaning that our tangent line is another -90° off from the hypotenuse. Finally, add the two numbers together and you have your answer! $(-30)°+(-90)° = \boxed{(-120)°}$