Question

(10 points) Suppose that the position of one particle at time t is given by x1 = 3 sin(t), yı = 2 cos(t), 0 <t<21 and the position of a second particle is given by x2 = -3 + cost), y2 = 1 + sin(t), 0313 21. (a) How many points of intersection are there for these paths? Number of intersection points = 2 (b) How many of these points of intersection are collision points? In other words, how many times are the particles in the same place at the same time? Number of collision points = 1 (c) List the collision points. List them in terms of increasing X-coordinates. If several collision points have the same x-coordinate, list those in order of increasing y-coordinates. Type an upper-case "N" in each unused answer blank. First collision point: (x, y) = (-3 0 ) Second collision point: (x, y) = ( 3 11 Third collision point: (x, y) =( 3 (d) Now suppose that the path of the second particle is given by x2 = 3 + cos(t), yz = 1+ sin(t), 0<t<21 and the path of the first particle remains the same. How many points of intersection are there for these paths? Number of intersection points = 2 (e) For the situation described in (d), how many of these points of intersection are collision points? Number of collision points = 1

Answer 1

Given position of particle one at time t as -21 = 3sin(t) ,    Y1 = 2cos(t) 0<t < 27 .:. sinº(t) +cos? (t) = 3)2+(2) as   sind(t) + cos' (t) = 1    al = 1 9 4 This equation represents equation of the ellipse. Position of particle two at time t as 12 = -3 + cos(t) ,    y2 = 1 + sin(t) 0<t < 27 I-TA (1) us       cos(t) = 12 +3 :. sinº(t) + cos²(t) = (y2 – 1)2 + (22+3)2 = (y2 – 1)+ (22+3)2 = 1 This equation represents equation of the circle with unit radius. 0 X We can see path taken by particle one and particle two in the above graph.

a. When two paths intersect, at the point of intersection 11 12 and   Y1 Y2 . But the time of intersection need not be same. (-2.097, 1.43) 0 X (-3,0) From graph we can see that there are two points of intersection at (-3.0) and (-2.097, 1.43) . $\therefore$ Number of intersection points= 2.

b. The collision points are the points of intersection when   11 12 and   Y1 Y2 at same time. This means that not only the path intersects but both particles meet. $\therefore$ 11 12 $\Rightarrow$ 3sin(t) = -3+ cos(t)   $\rightarrow (1)$ Y1 Y2 ► 2cos(t) = 1 + sin(t) (2) Multipy equation (1) by 2. Add it to equation   (2) → 6sin(t) + 2cos(t) = -6 + 2cos(t) +1+ sin(t) 5sin(t) = -5 sin(t) -1 t= sin(-1) 3л = t = . 2 $\therefore$ At   , particles collide. 3 2 Substitute in    Ir and  $y_{1}$ to get the collision point at time 3 2 . 37 11 3sin 2    C1 = -3=22 У1 — 3 2cos(- 2 Th = 0 = y2 The collision point is (-3.01. $\therefore$ Number of collision point =1.

c. The collision point at time 3 2 is (-3.01. First collision point: (x,y) = (-3.01. Second collision point: (x,y) = (N,N). Third collision point: (x,y) = (N,N).

d. Now position of particle two at time t as 12 = 3 + cos(t) ,    y2 = 1 + sin(t) 0<t < 27 I-TA (1) us , $cos(t)=x_{2}-3$ (42 – 1)2 + (22 - 3)2 = This equation represents equation of the circle with unit radius. (2.097, 1.43) 0 X (3,0) From graph we see that there are two points of intersection at   and . $\therefore$ Number of intersection points= 2.

e. For collision point consider $\therefore$ 11 12 $\Rightarrow$$3sin(t)=3+cos(t)$  $\rightarrow (3)$ Y1 Y2 ► 2cos(t) = 1 + sin(t) $\rightarrow (4)$ Multipy equation by 2. Add it to equation   But   $\therefore$ There is no point of collision. $\therefore$ Number of collision point =0.