Puzzles arithmetic – Can you Avoid the Spear-Wielding Gladiator? Team FunTrove - October 2, 2022 0 [ad_1] arithmetic – Can you Avoid the Spear-Wielding Gladiator? – Puzzling Stack Exchange Stack Exchange Network Stack Exchange community consists of 182 Q&A communities together with Stack Overflow, the most important, most trusted on-line group for builders to study, share their information, and construct their careers. Visit Stack Exchange Log in Sign up Puzzling Stack Exchange is a query and reply web site for many who create, clear up, and research puzzles. It solely takes a minute to enroll. Sign as much as be a part of this group Anybody can ask a query Anybody can reply The finest solutions are voted up and rise to the highest Asked at this time Viewed 14 instances $begingroup$ You are trapped in a round coliseum, and a gladiator with a spear is chasing you. You cannot defend your self, however you may run quicker than the gladiator. You run at 11 ft per second, and the gladiator runs at 10 ft per second. The gladiator’s spear has a spread of 10 ft. You should keep at the very least 10 ft away from the gladiator, or else you may be killed. The gladiator is just not sensible sufficient to take a strategic path that optimally chases you down. They solely head straight in your path. You management the beginning place of your self and the gladiator. How large does the coliseum should be for you to have the ability to keep away from the gladiator indefinitely? Generalized, in case your velocity is V, the gladiator’s velocity is W, and the spear size is X, how large does the coliseum should be? requested 53 minutes in the past Jonathan is a brand new contributor to this web site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. $endgroup$ 2 $begingroup$ Normalise models so the gladiator runs at velocity $1$, you run at velocity $s>1$ and the spear has size $d$. The thought is to have the gladiator $G$ run in a circle $Gamma_1$ of some radius $r$ centred within the coliseum’s origin $O$. You, $Y$, put your self on the finish of the vector of size $d$ pointing away from $G$ within the path of his instantaneous velocity (tangent to $Gamma_1$) and run in a bigger circle $Gamma_2$ concentric with $Gamma_1$. Then $triangle OGY$ is right-angled at $G$ with $OG=r,GY=d$ and therefore $OY=sqrt{r^2+d^2}$. You wish to keep the right-angled triangle’s form and maintain it spinning round $O$; since velocity is proportional to radius, to match you and $G$‘s most speeds you want$$frac{sqrt{r^2+d^2}}r=sqrt{1+(d/r)^2}=s$$ $$r=frac d{sqrt{s^2-1}}$$and a coliseum of dimension $rs$. For the numerical values right here with $s=1.1$ and $d=1$ this provides $r=frac{10}{sqrt{21}}=2.1821789dots$ ($21.821dots$ ft) and the coliseum radius as $frac{11}{sqrt{21}}=2.4003967dots$ ($24.003dots$ ft). answered 11 minutes in the past Parcly TaxelParcly Taxel 4,07344 silver badges3737 bronze badges $endgroup$ Your privateness By clicking “Accept all cookies”, you agree Stack Exchange can retailer cookies in your machine and disclose data in accordance with our Cookie Policy. Accept all cookies Customize settings [ad_2] LEAVE A REPLY Cancel reply Please enter your comment! Please enter your name here You have entered an incorrect email address! Please enter your email address here Save my name, email, and website in this browser for the next time I comment.