Kinematics: 100 meter dash when one runner starts 3 seconds later

[jnuz73hbn]

Homework Statement

Hello, I think I have a simple error in my thinking. Topic 100m sprint: Man A and man B have to sprint 100m. Man v(a)= 5m/s . Man B starts 3s later his speed v(b)= 7m/s. Who wins? Graphically I have arrived at the following functions, are they correct- with "t-3"?

Relevant Equations

$$s_B(t) = 7 \cdot (t - 3) + 0$$

$$t_A= 3s$$
$$ s= 100m$$
$$v_A= 5m/s$$
$$v_B=7m/s$$
$$s(t)=7(t-3)+0$$

 

[berkeman]

Homework Statement: Hello, I think I have a simple error in my thinking. Topic 100m sprint: Man A and man B have to sprint 100m. Man v(a)= 5m/s . Man B starts 3s later his speed v(b)= 7m/s. Who wins? Graphically I have arrived at the following functions, are they correct- with "t-3"?
Relevant Equations: $$s_B(t) = 7 \cdot (t - 3) + 0$$

$$t_A= 3s$$
$$ s= 100m$$
$$v_A= 5m/s$$
$$v_B=7m/s$$
$$s(t)=7(t-3)+0$$

I'm having trouble following your work. Can you say in words what you are doing with those equations?

I would find the time when the two runner positions are equal, and then find how far they are at that time. That would tell me if the 2nd runner catches the lead runner...

 

[kuruman]

There is only one function that you have and that is $s(t)=7(t-3)+0.$ That is presumably the position of runner B as a function of time. The rest of your equations are the given quantities. What is your strategy for answering the question? @berkeman gave you one. Here is my hint: When B starts the race, how long will it take each runner to reach the finish line?

 

[Orodruin]

Here's a hint at a different approach from those suggested:
When will A have completed the race? How far will B have come when that happens? If B has come more than 100 m then B has already completed the race before A.

I find this approach somewhat easier to handle arithmetically as you don't need to compute the time when B crosses the finish line. You only need to deal with multiplication of integers at worst.

 

[PeroK]

Just don't use Zeno's method.

 

[PeroK]

My approach would be to calculate the times for A and B to run 100m and see how much shorter B's time is. That tells you everything you need to know.

 

[jnuz73hbn]

There is only one function that you have and that is $s(t)=7(t-3)+0.$ That is presumably the position of runner B as a function of time. The rest of your equations are the given quantities. What is your strategy for answering the question? @berkeman gave you one. Here is my hint: When B starts the race, how long will it take each runner to reach the finish line?

My only question is whether I have set up this equation for B correctly, whether I should write the 3 second delay in the equation as $$s(t)=7(t+3)$$ or $$s(t)=7(t-3)$$

 

[jnuz73hbn]

I'm having trouble following your work. Can you say in words what you are doing with those equations?

I would find the time when the two runner positions are equal, and then find how far they are at that time. That would tell me if the 2nd runner catches the lead runner...

My only question is whether I have set up this equation for B correctly, whether I should write the 3 second delay in the equation as $$s(t)=7(t-3)$$ or $$s(t)=7(t+3)$$

 

[PeroK]

My only question is whether I have set up this equation for B correctly, whether I should write the 3 second delay in the equation as $$s(t)=7(t+3)$$ or $$s(t)=7(t-3)$$

It depends how you define $s(t)$. If there are two runners, there are two displacements. I would use $s_A(t)$ and $s_B(t)$.

 

[Orodruin]

My approach would be to calculate the times for A and B to run 100m and see how much shorter B's time is. That tells you everything you need to know.

This is my backbone reaction too, but then I realized it requires dividing 100 by 7 (sure, not that difficult if you remember that 7^2 = 49). Since 100/5 is an easy integer, that integer minus 3 times 7 is straightforward.

 

[PeroK]

My only question is whether I have set up this equation for B correctly, whether I should write the 3 second delay in the equation as $$s(t)=7(t+3)$$ or $$s(t)=7(t-3)$$

Try computing $s(0)$ or $s(3)$.

 

[kuruman]

My only question is whether I have set up this equation for B correctly, whether I should write the 3 second delay in the equation as $$s(t)=7(t+3)$$ or $$s(t)=7(t-3)$$

Here is how I would think this through. Imagine that each runner wears a smartwatch that starts the stopwatch function when the wearer starts moving. The positions of the runners will be given by
$s_A(t_A)=5~t_A$ and $s_B(t_B)=7~t_B$. We want to write expressions involving a single clock time $t$ which means that we have to choose one of the two possibilities. Before choosing, we note that B's stopwatch display shows 3 s less than A's display: $t_B=t_A-3~$s.

Therefore, if we choose $t_A=t$ as the common time, we write $s_A(t)=5~t$ and $s_B(t)=7~(t-3).$
If we choose $t_B=t$ as the common time, we write $s_A(t)=5~(t+3)$ and $s_B(t)=7~t.$

 

[kuruman]

Just don't use Zeno's method.

Why not? I think that Zeno's method will work if you ask the right person.

If you ask a race spectator, he will give the standard reply, something like "B starts 15 m behind A. In the time B covers the 15m, A will have advanced an additional ~10 m and by the time B covers that, A will have advanced $\dots$ and so on ad infinitum."

However, if you ask one of the runners, say A, he will say, "I see B 15 m behind me approaching me at 2 m/s. He will cover that distance in $\frac{15}{2}~$s. I also see the finish line 85 m ahead of me approaching me at 5 m/s. It will cover that distance in $\frac{85}{5}~$s.

If $\frac{15}{2}<\frac{85}{2}$, I lose the race;
Else if $\frac{15}{2}>\frac{85}{5}$, I win the race;
Else it's a tie;
End If"

Similar considerations apply to runner B who sees runner A and the finish line approaching him from the same direction with respective speeds 2 m/s and 7 m/s and from respective initial distances 15 m and 100 m. It is interesting to note that, as far as resolution of Zeno's paradox is concerned, we apparently we have two preferred inertial frames.

 

[PeroK]

Why not? I think that Zeno's method will work if you ask the right person.

If you ask a race spectator, he will give the standard reply, something like "B starts 15 m behind A. In the time B covers the 15m, A will have advanced an additional ~10 m and by the time B covers that, A will have advanced $\dots$ and so on ad infinitum."

However, if you ask one of the runners, say A, he will say, "I see B 15 m behind me approaching me at 2 m/s. He will cover that distance in $\frac{15}{2}~$s. I also see the finish line 85 m ahead of me approaching me at 5 m/s. It will cover that distance in $\frac{85}{5}~$s.

If $\frac{15}{2}<\frac{85}{2}$, I lose the race;
Else if $\frac{15}{2}>\frac{85}{5}$, I win the race;
Else it's a tie;
End If"

Similar considerations apply to runner B who sees runner A and the finish line approaching him from the same direction with respective speeds 2 m/s and 7 m/s and from respective initial distances 15 m and 100 m. It is interesting to note that, as far as resolution of Zeno's paradox is concerned, we apparently we have two preferred inertial frames.

If Zeno had known about changing reference frames he would have solved his paradox himself.

 

[kuruman]

If Zeno had known about changing reference frames he would have solved his paradox himself.

Galileo had a chance, but I guess he didn't know about it.