Jack and Jill may be skating still

First, we determine the distance both skaters are from the edge of the rink when they meet each other. To do this, it is easiest to make a table to organize our work. We are given the speeds at which Jack and Jill skate, so letting $t_m$ be the time they meet and plugging in the values into the table gives us the following table:

$\begin{array}{|c|c|c|c|}\hline & \text{Rate} & \text{Time} & \text{Distance} \\ \hline \text{Jill} & 3 & t_m & 3t_m \\ \hline \text{Jack} & 5 & t_m-3 & 5(t_m-3) \\ \hline\end{array}$

Since the time it takes for both Jack and Jill to travel this distance are the same, we have $3t_m=5(t_m-3)\implies t_m=7.5.$

Therefore the two skaters meet at time $t=7.5$ seconds, and when they do meet they are at a distance of $3(7.5)=22.5$ meters away from the original end of the rink. This implies that Jill travels the remaining $50-22.5=27.5$ meters scooped up in Jack's arms.

Now, the Law of Conservation of Momentum comes into play. When Jill is skating, she exerts a momentum of $(50\text{ kg})(3 \text{ m/s})=150\text{ N}\cdot\text{s}$ , and when Jack is skating he exerts a momentum of $(80\text{ kg})(5 \text{ m/s})=400\text{ N}\cdot\text{s}$ . Therefore, when they "collide", the momentum of the system is therefore $550\text{ N}\cdot\text{s}$ . This does not change once Jill gets scooped up in Jack's arms. Since the two have a combined mass of $130\text{ kg}$ , their momentum is $130v\text{ N}\cdot\text{s}$ , where $v$ is the speed of the couple. This means that $550=130v\implies v=\dfrac{55}{13}=4.23$ . Thus, combining this with our earlier work, we get that the time it takes for the couple to travel to the end of the ice rink is $\dfrac{\text{distance}}{\text{rate}}=\dfrac{27.5\text{ m}}{4.23\text{ m/s}}=6.5\text{ s}.$ Finally, we have that they reach the end of the ice rink at time $t=6.5\text{ s}+7.5\text{ s}=\boxed{14\text{ s}}.$

The first step is to find the position of Jack and Jill when they meet. To do this, we find the intersection of the lines $y = 3t + 9$ and $y = 5t$ where $y$ is position in meters and $t$ is time in seconds. We get $(t,y) = (4.5$ seconds $, 22.5$ meters $)$ . Because this is the time after Jack began skating, we must add $3$ seconds to it, which gives us $7.5$ seconds.

When Jack picks Jill up, their combined velocity changes and, because there is no friction on the ice to slow them down, they will continue at this velocity for the rest of their path across the rink. We can calculate their velocity using conservation of momentum. That is, $m_{Jack}\cdot{v_{Jack}} + m_{Jill}\cdot{v_{Jill}} =$ $m_{Jack and Jill}\cdot{v_{Jack and Jill}}$ where $m_i$ and $v_i$ represent the mass and velocity of person(s) $i$ . Substituting in the given values to this equation, we get $80\cdot5 + 50\cdot3 = (80 + 50)\cdot{v_{Jack and Jill}} \to$ $v_{Jack and Jill} = \frac{550}{130} = \frac{55}{13}$ . Because Jack and Jill travelled $22.5$ meters before meeting, they have $27.5$ meters more to travel, so the time it takes them to reach the end is $\frac{27.5}{\frac{55}{13}} = 6.5$ seconds. Adding this to the time before their meeting, $7.5$ seconds, gives us a final answer of $14$ seconds.

First, Jack scoops up Jill at $t_1$ , where $3 \times 3+3t_1=5_1 \Rightarrow t_1=7.5$ seconds. After that, their new velocity by the conservation of momentum is $v$ , where $50 \times 3 + 80 \times 5=v(50+80) \Rightarrow v=\frac{55}{13}$ . Therefore the time taken is $7.5+\frac{50-7.5 \times 3}{\frac{55}{13}}=14$ seconds.

We know that Jill starts to skate first, at t=0 seconds, and then Jack follows her after 3 seconds. And then in some point they meet and Jack picks her up and they skate together for the remaining distance.

So, we have to find the time and the distance at which they meet. The time is with respect to Jill's starting time. We know that that they will meet up after some distance because Jack is moving much faster than Jill although she starts skating first. Let t_Jill be Jill's skating time,

        d_ Jill = d_ Jack

and by using (distance = speed x time), v_ Jill(t_ Jill)=v_ Jack(t_ Jack)

           v_ Jill(t_ Jill)=v_ Jack(t_ Jill - 3)

Plug in the numbers we get,

           (3 m/s)(t_Jill)=(5 m/s)(t_Jill - 3)

            3 t_Jill=5 t_Jill - 15

            2 t_Jill= 15

               t_Jill= 7.5 seconds

The distance at which they meet ,

             v_Jill x t_Jill

            =(3 m/s)(7.5s)

            =22.5 meters

The distance left for both of them to cover,

           (50 - 22.5) meters

           =27.5 meters

At this point Jack scoops Jill up and they move together. This situation is similar to an inelastic collision.

Ignoring drag or friction on the ice, we can assume that total momentum is conserved, and we can use this formula

italic (m 1)(u 1)+(m 2)(u 2)=(m 1+m 2)(v)

Plug in the numbers, we get

        (50)(3) + (80)(5) = (50+80)v

        550=(130)v

        v= 4.23 m/s [speed of Jack  and Jill moving together]

Time of both of them to cover the remaining 27.5 meters

= (27.5)/(4.23)

=6.5 seconds

This is not the time where they reach the other side of the rink. This is just the time for the 27.5 meters

Thus, the time where they reach the other side of the rink

=6.5 seconds + (the time where Jack and Jill met)

=6.5 seconds + 7.5 seconds

=14 seconds

For jill d=vt=3*3=9 m Jack travels 9+3t distance before scopping up.....velocity of jack is 5 m/s...9+3t=5t t=4.5sec

Total time=3+4.5=>7.5... Total distance travvelled uptill now...3*7.5=22.5...remaining distance =50-22.5=>27.5 Now conserve momentum...and get v final(velocity after scooping) den divide it with 27.5 as t=d/v...der total time =t1+t2

Jill: d〓vt〓3m/s x 3s〓9m Jack 〓 Jill: 9m + 3m/s t 〓5m/s t; 2m/s t 〓 9m; t〓4.5s Total time so far 7.5s Total distance so far 7.5s x 3m/s 〓 22.5 Distance remaining: 50 - 22.5 〓 27.5m New velocity after scoop: m1v1 + m2v2 〓 m3v3; 50x3 + 80x5 〓 m3v3 ,130v3 〓 550; v3〓4.23m/s Remaining time: 27.5 m / 4.23 m/s 〓 6.5s Total time〓6.5s + 7.5s 〓14s

Jill: d〓vt〓3m/s x 3s〓9m Jack 〓 Jill: 9m + 3m/s t 〓5m/s t; 2m/s t 〓 9m; t〓4.5s Total time so far 7.5s Total distance so far 7.5s x 3m/s 〓 22.5 Distance remaining: 50 - 22.5 〓 27.5m New velocity after scoop: m1v1 + m2v2 〓 m3v3; 50x3 + 80x5 〓 m3v3 ,130v3 〓 550; v3〓4.23m/s Remaining time: 27.5 m / 4.23 m/s 〓 6.5s Total time〓6.5s + 7.5s 〓14s

Assume Jack is A and Jill is I, so mI = 50 kg mA = 80 kg vI = 3 m/s vA = 5 m/s tA = tI - 3 s

first step, calculate when Jack meet Jill and scoops up her, when they have traveled same distance: sA = sI vA . tA = vI . tI 5 . (tI - 3) = 3 . tI so tI = 7.5 s and sI = 22.5 m

second step, calculate their combined velocity using energy conservation: EkA + EkI = EkC 0.5 . mA . vA^2 + 0.5 . mI . vI^2 = 0.5 . mC . vC^2 0.5 . 80 . 5^2 + 0.5 . 50 . 3^2 = 0.5 . (80+50) . vC^2 vC^2 = 245/13 so vC = 4.3412 m/s

third step, calculate time for the remaining distance: sR = 50 - sI = 50 - 22.5 = 27.5 m tR = sR / vC = 27.5 / 4.3412 = 6.3346 s

fourth step, calculate the total time: t = 7.5 + 6.3346 = 13.8346 s

t = t' + t'' t' = time needed for Jack to meet Jill t'' = time needed for those 2 reach the other side after meet t' = 3 * 3 / (5-3) = 4.5 seconds

use momentum v' = (50*3 + 80 *5)/130 t'' = 41 *130 / 550 = 14.2