0
$\begingroup$

At the same moment two particles start respectively from vertices $B$ and $C$ of triangle $ABC$ which has a right angle at $C$. The particles move at constant speeds and arrive at vertex $A$ at the same moment.

If the time of travel is $(AB/AC)^{3/2}$ and the length of $BC$ is equal to the sum of the constant speeds, determine the time of travel.

$\endgroup$
0
1
$\begingroup$

let $\tau = (\frac{AB}{AC})^{3/2}$ and $BC=v_1+v_2$
We know that $\tau_1=\tau_2=\tau$ $\rightarrow$ $\frac{AB}{v_1}=\frac{AC}{v_2}$ $$\frac{v_1}{v_2}=\frac{AB}{AC}=\tau^{2/3}\hspace{1 pt}...(1)$$
then $AB=v_1.\tau$ and $AC=v_2.\tau$ $$\sqrt{AB^2-AC^2}=v_1+v_2$$ $$\tau.\sqrt{v_1^2-v_2^2}=v_1+v_2$$ square both sides $\rightarrow \tau^2.(v_1-v_2)=v_1+v_2$ $$v_1(\tau^2-1)=v_2(\tau^2+1)$$ $$\frac{v_1}{v_2}=\frac{\tau^2+1}{\tau^2-1}$$ from the (1) we know that : $$\frac{\tau^2+1}{\tau^2-1}=\tau^{2/3}$$ arrange a little bit : $$\tau^{8/3} -\tau^2-\tau^{2/3}-1=0 \hspace{2 pt}...(2)$$ set $\tau^{2/3}=p$, and eq (2) become $p^4-p^3-p-1=0$, factor and we get : $(p^2+1)(p^2-p-1)=0$
clearly $p \neq \pm i $ . And from the other , we get $p=\frac{1}{2}(1 \pm \sqrt{5})$. Because $\sqrt{5} > 1$ , the only solution for $p$ is :

$$p=\frac{1}{2}.(1+\sqrt{5})$$ $$\tau=(\frac{1}{2}.(1+\sqrt{5}))^{3/2} \approx 2.05817$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.