Two cars start at the same time from the junction of two roads one on each road with the uniform speed of $v m/sec$.

Question

Two cars start at the same time from the junction of two roads one on each road with the uniform speed of $v m/sec$. If the roads are inclined at $120°$, show that the distance between them increases at the rate of $\sqrt {3}v m/s$

My Approach:

Given: $\dfrac {dx}{dt}=v m/s$ and $\dfrac {dy}{dt}=v m/s$
Using cosine law, $$s^2=x^2+y^2-2xy\cos (120°)$$ $$s^2=x^2+y^2+xy$$ differentiating both sides w.r.t $t$ $$2s.\dfrac {ds}{dt}=2x.\dfrac {dx}{dt}+2y.\dfrac {dy}{dt}+x.\dfrac {dy}{dt}+y.\dfrac {dx}{dt}$$ $$2s\dfrac {ds}{dt}=3xv+3yv$$ $$2s.\dfrac {ds}{dt}=3v(x+y)$$

How do I proceed further?

Answer 1

Let $S=vt$ be the distance each car passes. Then: $\frac{dS}{dt}=v$. The distance between the cars is: $$D=\sqrt{2S^2-2S^2\cos 120^\circ}.$$ Differentiating the distance formula: $$\frac{dD}{dt}=\frac{4S\cdot \frac{dS}{dt}+2S\cdot \frac{dS}{dt}}{2\sqrt{2S^2+S^2}}=\sqrt{3}v.$$

Answer 2

I think $x$ and $y$ are used in a confusing way here

Let one road coincide with OX and the other be at $\frac{\pi}{3}$ in a positive direction, then

for the first car we have $$(x_1(t), y_1(t)) = (vt, 0)$$ and for the second car $$(x_2(t),y_2(t)) = \left(-vt\cdot\cos\left(\frac{\pi}{3}\right), vt\cdot\sin\left(\frac{\pi}{3}\right)\right)$$

now $$s^2(t) = (x_1(t)-x_2(t))^2 + (y_1(t) - y_2(t))^2$$

$$\frac{\mathrm{d}}{\mathrm{d}t}(s^2(t)) = \frac{\mathrm{d}}{\mathrm{d}t}\left(v^2t^2\cdot\left(\left(1+\cos\left(\frac{\pi}{3}\right)\right)^2+\sin^2\left(\frac{\pi}{3}\right)\right)\right)$$

which is $$\frac{\mathrm{d}}{\mathrm{d}t}(s^2(t)) = \frac{\mathrm{d}}{\mathrm{d}t}\left(3v^2t^2\right)$$

Answer 3

Just use trigonometry. We have a triangle with two sides length v and an angle of 120 degrees between them. The 3rd side is the distance between the cars, say = d Drop perpendicular from the 120 degree angle to form a right angle triangle, so the split angle becomes 60 degrees. Standard ration for sin 60 is root3 over 2 Therefore sin 60 = root3 over 2 = d over 2v Hence d = root3 times v

Had not worked out how to use the symbology here so had to use the above description.