Three amigos

Let $u=1 m/s$ be Ursula's velocity and $v=2 m/s$ Sven and Margo's velocity. First lets compute $t_m$ . In order for Margo to meet Ursula in the shortest amount of time, she must move on a straight line towards their pre-calculated point of intersection. Let this point be $(0,x)$ , then $x=ut_m$ and Margo will travel the distance of $vt_m$ . Also from the right triangle formed by the points $(100, 0), (0, 0)$ and $(0,x)$ , using Pythagorean Theorem we obtain $100^2+(ut_m)^2=(vt_m)^2$ . Now solving a simple equation we get $t_m=\frac{100}{\sqrt{v^2-u^2}}\approx 57,74s$ . Now lets move onto $t_s$ . Assume that the distance between Sven and Ursula at a given time is $l$ . The angle between $\vec{u}$ and $\vec{v}$ is $\varphi$ and it is time-dependent. We have $-\frac{dl}{dt}=v-u\cos\varphi$ . Integrating, we get $l=\int^{t_s}_{0}(v-u\cos\varphi)dt$ . Also, since Sven and Ursula travel the same distance across the $Y$ axis, we get $y=ut_s=\int^{t_s}_{0}u\cos\varphi dt$ . Combining the last two equations, we conclude that $t_s=\frac{100v}{v^2-u^2}\approx 66,67 s$ .

Therefore the final answer is $t_s-t_m=8,93s$ .

Margo takes the shortest possible path, so her path is a straight line towards a point on the $y$ -axis, where she will meet Ursula. To calculate this point, we can note that after $t_m$ seconds, Ursula walked $t_m$ metres while Margo walked $2t_m$ metres. Applying Pythagora's theorem, we have $t_m^2 + 100^2 = (2t_m)^2$ . This gives $t_m = \frac{100}{\sqrt{3}}$ seconds.

Sven, on the other hand, will walk directly towards Ursula. Let the angle made between their directions of motion be $\theta$ . Considering the distance moved in the positive $y$ -direction, we have $\int_0^{t_s} 2 \cos \theta \,\mathrm{d}t = t_s \Rightarrow \int_0^{t_s} \cos \theta \,\mathrm{d}t = \frac{t_s}{2}$ . Considering the straight-line distance between them (by considering the velocity of approach from Sven to Ursula), we have $\int_0^{t_s} 2 - \cos \theta \,\mathrm{d}t = 100 \Rightarrow 2t_s - \frac{t_s}{2} = 100$ . This gives $t_s = \frac{200}{3}$ seconds.

So $t_s - t_m = \frac{200}{3} - \frac{100}{\sqrt{3}} = 8.932$ seconds.

The distance covered by Margo is easily seen to be $\frac{100}{\sqrt{3}}$ since she travels in a straight line to meet Ursula on the y axis.

The trajectory followed by Sven is more difficult to calculate since it is given by the differential equation $\frac{d\mathbf{r}}{dt}=2\frac{(0,t)-\mathbf{r}}{|(0,t)-\mathbf{r}|}$ with initial condition $\mathbf{r}(0)=(100,0)$

Instead of solving this analytically, we can try a numerical method since only three digits of precision are required. The code used for the same was as follows

x,y,t=100.,0.,0. dt=1e-5 while x>1e-6: dx=-x dy=t-y norm=sqrt(dx dx+dy dy) x+=2 dt dx/norm y+=2 dt dy/norm t+=dt print t

Let u be Ursula's speed and v-Sven's speed. First find tm for Margo to meet ursula in the shortest amount of time, she should move in a straight line to their point of intersection. Assume this point is located at (0; x), then x = u tm and Margo travels the distance v *tm. And from Pitagora in triangle (100;0);(0;0);(0; x) we obtain 100^2+(utm)^2 = (vtm)^2. It results that tm = 100/sqrt(v^2-u^2) = 57,74s. Now lets compute ts. Let the distance betwen Sven and Ursula at a given time be l. The angle betweem u(vectorial) and v(vectorial) is alpha and it's time-dependant. We have -dl/dt = v -u cos(alpha). Integrating, we obtain l =(Integral) from ts to 0 (v-u cos alpha)dt. But because Sven and Ursula travel the same distance across the Y axis, we get y = u ts =Integral from ts to 0 u*cosalpha dt. From this + previous equation, we get that ts = 100v/(v^2-u^2)= 66,67s. Answer ts -tm = 8,93s.

If we drawn this problem, we will get path of margo linier, and path of sven quarter of elips: Sven meet with ursula after ursula walk distance $x m$ for time $t_{s}$ with use aproximation of perimeter of elips: http://www.mathsisfun.com/geometry/ellipse-perimeter.html $t_{s}=\frac{\pi\sqrt{(\frac{100^2+x^2}{2})}}{4}=x$ , so that $t_s=x=66.781$ And also: Margo meet with ursula for distance $y$ , and time $t_m$ $t_{m}=t_{m}=\sqrt{\frac{100^2+y^2}{2}}=y$ $t_{m}=y=57.735$ and from the problem $t_{s}-t_{m}=66.781-57.735=9.046$ . And we done.

Let point O is the origin , A(100 , 0) and margo meet ursula at B(0 , y_m) . in order to walk in the shortest amount of time Margo must walk straight ahead to point B. Since Margo’s speed twice than Ursula’s , then AB = 2 ym and OAB = 30°

AB = OA /cos30°

AB = 200/ √3

t_m = (200/√3 m) / (2m/s) = 57.74 s

Sven always walks directly towards Ursula. Since Sven’s speed twice

than Ursula’s , thus Sven’s track is an elliptical with semi-major axis =

100 and semi-minor axis = y_s

Lenght of AC = ¼ x circumference of an ellipse

AC = 2 y_s

¼ (2π √( (100^2 + ys^2)/2 ) = 2 ys

y_s = 66.78

AC = 2 x 66.78 = 115.47 m

t_s = (115.47 m) / (2m/s) = 66.78 s

t s – t m = 66.78 s - 57.74 s = 9.04 s

Let us first do the easier calculation of $t_m$ . If Margo walks to meet Ursula in the shortest time, then she is also walking along the path of shortest distance, i.e. a straight line. We can therefore construct the following equation (the Pythagorean theorem for a triangle with legs of length $1t_m$ meters and $100$ meters and hypotenuse of length $2t_m$ meters),

$t_m^2+100^2=(2t_m)^2$

which allows us to solve for $t_m=57.735~\mbox{s}$ .

Now for Sven's path. This is a classic case of a pursuit curve, which you can read all about here . For the initial conditions and velocities specified, the intersection point is given by $(0~\mbox{m}, 2 \times 100/(2^2-1)=66.667~\mbox{m}).$ Therefore $t_s=66.667~\mbox{s}$ . Hence Margo reaches Ursula $8.935~\mbox{s}$ before Sven.