Margo and Sven are outside in a field, standing right next to each other and talking. We'll put x,y coordinates on the field and measure everything in meters. Ursula, a friend of both Margo and Sven, is also outside walking due north (positive y-direction) at 1 m/s . When Ursula crosses the origin (at t = 0 ) Margo and Sven (at coordinates ( 1 0 0 , 0 ) ) notice her and decide to walk over and say hello.
Margo knows a little physics, and so she figures out how she should walk at 2 m/s to meet Ursula in the shortest amount of time. Sven doesn't know any physics, so he just always walks directly towards Ursula, also at 2 m/s , until he catches up to her.
If the time Sven meets Ursula is t s and the time Margo meets Ursula is t m , what is t s − t m in seconds?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
This problem is a modified version of a problem in I E Irodov's book "Problems in General Physics"
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 2 t m metres. Applying Pythagora's theorem, we have t m 2 + 1 0 0 2 = ( 2 t m ) 2 . This gives t m = 3 1 0 0 seconds.
Sven, on the other hand, will walk directly towards Ursula. Let the angle made between their directions of motion be θ . Considering the distance moved in the positive y -direction, we have ∫ 0 t s 2 cos θ d t = t s ⇒ ∫ 0 t s cos θ d t = 2 t s . Considering the straight-line distance between them (by considering the velocity of approach from Sven to Ursula), we have ∫ 0 t s 2 − cos θ d t = 1 0 0 ⇒ 2 t s − 2 t s = 1 0 0 . This gives t s = 3 2 0 0 seconds.
So t s − t m = 3 2 0 0 − 3 1 0 0 = 8 . 9 3 2 seconds.
The distance covered by Margo is easily seen to be 3 1 0 0 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 d t d r = 2 ∣ ( 0 , t ) − r ∣ ( 0 , t ) − r with initial condition r ( 0 ) = ( 1 0 0 , 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 = 4 π ( 2 1 0 0 2 + x 2 ) = x , so that t s = x = 6 6 . 7 8 1 And also: Margo meet with ursula for distance y , and time t m t m = t m = 2 1 0 0 2 + y 2 = y t m = y = 5 7 . 7 3 5 and from the problem t s − t m = 6 6 . 7 8 1 − 5 7 . 7 3 5 = 9 . 0 4 6 . 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 1 t m meters and 1 0 0 meters and hypotenuse of length 2 t m meters),
t m 2 + 1 0 0 2 = ( 2 t m ) 2
which allows us to solve for t m = 5 7 . 7 3 5 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 m , 2 × 1 0 0 / ( 2 2 − 1 ) = 6 6 . 6 6 7 m ) . Therefore t s = 6 6 . 6 6 7 s . Hence Margo reaches Ursula 8 . 9 3 5 s before Sven.
Problem Loading...
Note Loading...
Set Loading...
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 = u t m and Margo will travel the distance of v t m . Also from the right triangle formed by the points ( 1 0 0 , 0 ) , ( 0 , 0 ) and ( 0 , x ) , using Pythagorean Theorem we obtain 1 0 0 2 + ( u t m ) 2 = ( v t m ) 2 . Now solving a simple equation we get t m = v 2 − u 2 1 0 0 ≈ 5 7 , 7 4 s . Now lets move onto t s . Assume that the distance between Sven and Ursula at a given time is l . The angle between u and v is φ and it is time-dependent. We have − d t d l = v − u cos φ . Integrating, we get l = ∫ 0 t s ( v − u cos φ ) d t . Also, since Sven and Ursula travel the same distance across the Y axis, we get y = u t s = ∫ 0 t s u cos φ d t . Combining the last two equations, we conclude that t s = v 2 − u 2 1 0 0 v ≈ 6 6 , 6 7 s .
Therefore the final answer is t s − t m = 8 , 9 3 s .