How Long Must a Runner Accelerate to Finish a 10,000-m Run on Time?

[leroyjenkens]

Another hint: distance = velocity x time

I forgot about that one. I have d = (Vo + V)/2 times t as my distance formula. It uses average velocity. I was trying to replace just the V and keep the Vo/2, which was messing me all up, when I could have just replaced the whole thing with V since it's constant velocity which is the same as the average velocity.
So I got the distance during the constant velocity phase is d = 5.49t + .2t^2.

 

[Redbelly98]

Not quite, but you're getting closer. We can't just multiply the velocity by t, because we are already using t as the time for the acceleration phase. You need an expression for the time duration of the constant velocity phase, and multiply the velocity by that time.

Hint: he accelerates for a time t, and the total remaining time (at the start of the acceleration phase) is 3 minutes. So, the time duration of the constant velocity phase is _____?

 

[leroyjenkens]

Not quite, but you're getting closer. We can't just multiply the velocity by t, because we are already using t as the time for the acceleration phase. You need an expression for the time duration of the constant velocity phase, and multiply the velocity by that time.

Hint: he accelerates for a time t, and the total remaining time (at the start of the acceleration phase) is 3 minutes. So, the time duration of the constant velocity phase is _____?

I didn't even think about using the same variable for different values.
The time duration of the constant velocity phase is the total time minus the acceleration time. The acceleration time is .2t, so the time of the constant velocity phase is 180 - .2t.
Multiplying that times the velocity during the constant phase 5.49 + .2t, I get...
d = -.04t^2 + 34.902t + 988.2.
I hope I did that right.

 

[Redbelly98]

What does that give you for t when you solve it? You know the answer was t=3.1 s, so if you get that value then your thinking was correct.

 

[leroyjenkens]

What does that give you for t when you solve it? You know the answer was t=3.1 s, so if you get that value then your thinking was correct.

I got -27.45 and 900.
This would be the seconds during the constant velocity, right? So the time during acceleration would be 180 - 900 which doesn't make sense. I double checked my math and I got the right answer. I was supposed to use the quadratic formula, right?

 

[Redbelly98]

The acceleration time is .2t ...

No it isn't. Why would you think that?

Hint: he accelerates for a time t ...

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By the way, when is this problem due?

 

[leroyjenkens]

No it isn't. Why would you think that?

No idea. The acceleration time is t. I'll answer your question on post #37 again.

Hint: he accelerates for a time t, and the total remaining time (at the start of the acceleration phase) is 3 minutes. So, the time duration of the constant velocity phase is _____?

The acceleration time is t, and since there's 180 seconds left, the constant velocity phase would have 180 - t time.
If I put those two into the distance formula d = vt, then it gives me a binomial times a binomial which would multiply out into a quadratic equation. But since it's equal to d and not zero, I can't solve it as a quadratic. Quadratics always have to be equal to 0, right?

By the way, when is this problem due?

I'm not in this class yet, I think I'm going to take it next semester. I downloaded the book.

 

[Redbelly98]

The acceleration time is t.

Okay.

The acceleration time is t, and since there's 180 seconds left, the constant velocity phase would have 180 - t time.
If I put those two into the distance formula d = vt, then it gives me a binomial times a binomial which would multiply out into a quadratic equation. But since it's equal to d and not zero, I can't solve it as a quadratic.

That's correct about using 180-t. However we have the same issue with d that we had with t before! We are already using d to mean something else, so you need to express the distance during constant velocity some other way. Just like you found the time is 180-t instead of t during the constant velocity phase.

Quadratics always have to be equal to 0, right?

In order to solve them using the quadratic formula, yes.

I'm not in this class yet, I think I'm going to take it next semester. I downloaded the book.

Okay. Well, I have to say I wonder if you are really putting as much thought into this problem as you would if it were for a class you were taking. Sorry if I am wrong, but we are making extremely slow progress and that makes me wonder.

 

[leroyjenkens]

That's correct about using 180-t. However we have the same issue with d that we had with t before! We are already using d to mean something else, so you need to express the distance during constant velocity some other way. Just like you found the time is 180-t instead of t during the constant velocity phase.

Velocity during acceleration phase is v = 5.49 + .2t.
Distance during acceleration phase is d = 5.49t + .2t^2.
1100 is the total distance, so distance during constant phase is 1100 - 5.49t + .2t^2.

Okay. Well, I have to say I wonder if you are really putting as much thought into this problem as you would if it were for a class you were taking. Sorry if I am wrong, but we are making extremely slow progress and that makes me wonder.

I think this problem just has my number. I did all the others up to this one without having any help. This one is in the second chapter and it's ranked as a level 3 problem, which are "meant as challenges for the best students".

 

[Redbelly98]

Distance during acceleration phase is d = 5.49t + .2t^2.

No it isn't. You had it right earlier, but after our discussion has moved on to other phases of the problem, the earlier parts we worked out seem to get lost.

I think this problem just has my number. I did all the others up to this one without having any help. This one is in the second chapter and it's ranked as a level 3 problem, which are "meant as challenges for the best students".

That's understandable, it is indeed more advanced since the runner's motion consists of two parts.

Here is a suggestion to help with organizing everything, which is to make up a chart with all the variables in it. The chart consists of 3 parts: (A) the acceleration phase, (B) the constant velocity phase, and (C) the "net total" of (A) and (B). It would look something like this:

variable   (A)              (B)          (C)
--------   ---              ---          ---
time       t                180 - t      180
distance   5.49t + 0.1t^2                1100
v initial                                (not relevant)
v final                                  (not relevant)
accel.     0.2               0           (not relevant)

Note: time is in s, distance is in m

I have filled in some of the parts that we have figured out already. I will have to bow out of our discussion at this point, see if filling out this chart helps you organize your thoughts enough to solve it.

Good luck!

 

[leroyjenkens]

No it isn't. You had it right earlier, but after our discussion has moved on to other phases of the problem, the earlier parts we worked out seem to get lost.


That's understandable, it is indeed more advanced since the runner's motion consists of two parts.

Here is a suggestion to help with organizing everything, which is to make up a chart with all the variables in it. The chart consists of 3 parts: (A) the acceleration phase, (B) the constant velocity phase, and (C) the "net total" of (A) and (B). It would look something like this:

variable   (A)              (B)          (C)
--------   ---              ---          ---
time       t                180 - t      180
distance   5.49t + 0.1t^2                1100
v initial                                (not relevant)
v final                                  (not relevant)
accel.     0.2               0           (not relevant)

Note: time is in s, distance is in m

I have filled in some of the parts that we have figured out already. I will have to bow out of our discussion at this point, see if filling out this chart helps you organize your thoughts enough to solve it.

Good luck!

Well, I appreciate all the help and I'm surprised you stuck around this long. I don't view this as you giving up on me or anything, I just know you can't sit here forever on the same problem.
But this is the last response because I think I got it.
Here is how I set up the problem:

Total distance minus the constant distance which is 1100 - 5.49t + .1t^2
Using the distance equation I made that equal to the initial velocity of the constant phase which is 5.49 + .2t, times the time of the constant phase, which is 180 - t.

1100 - 5.49t + .1t^2 = (5.49 + .2t)(180 - t)

Solving for t, which is the time during the acceleration phase, I got 3.2. I think I didn't get 3.1 because of rounding I did earlier in the problem.