Implicit Trigonometry Differentiation Word Problem

Question

Two trains leave a station at the same time on tracks that are 60° to each other. One train travels at 80 km/h and the other travels at 100 km/h. At what rate are the trains separating 2 hours later?

I tried cosine law implicit differentiation but the answer is not even close to making sense. I'm not sure where my error lies.

Answer 1

I think the intention of this exercise is to use related rates and - as mentioned in the header - implicit differentiation.

So, if $d(t)$ denotes the distance between the trains at time $t$ and if $v_1,v_2$ denote the two velocities, you have according to the law of cosines

$$d^2(t) = (v_1t)^2+ (v_2t)^2 - 2v_1v_2t^2\cos 60° = (v_1^2+v_2^2-v_1v_2)t^2$$

Implicit differentiation gives

$$2d(t)\dot d(t) = 2(v_1^2+v_2^2-v_1v_2)t$$

or $$\dot d(t) = \frac{(v_1^2+v_2^2-v_1v_2)t}{d(t)} = \frac{(v_1^2+v_2^2-v_1v_2)t}{t\sqrt{v_1^2+v_2^2-v_1v_2}} = \sqrt{v_1^2+v_2^2-v_1v_2}$$

Answer 2

By the cosine law,

$$\frac d{dt}\sqrt{v_0^2t^2+v_1^2t^2-2v_0v_1t^2\cos\theta}.$$

Answer 3

Relative velocity is the vector difference of (uniform) velocity vectors.

$$v_{rel}=\sqrt{v_1^2+ v_2^2 - 2 v_1 v_2 \cos \gamma}$$ $$=\sqrt{100^2+ 80^2 - 100*80}$$

Answer 4

Solved it! I forgot to put in cos 60 before differentiating as it is an unchanging value. This makes the derivative massively easier to compute.

Answer: 91.65 km/h

Thanks for the feedback and help!

PS I'm not sure why I got down voted after I edited my post? Sorry, I'm new to this forum and not sure of the expectations for posting. :/