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

[leroyjenkens]

Homework Statement

A runner hopes to complete the 10,000-m run in less than 30.0 min. After exactly 27.0 min, there are still 1100 m to go. The runner must then accelerate at 0.20 m/s/s for how many seconds in order to achieve the desired time?

Homework Equations

Kinematic equations.

The Attempt at a Solution

Well I figured out at his current pace, he would finish the 1100 remaining meters in 200 seconds. However, he needs to complete it in 180 seconds. I'm not sure how to find out how long he has to accelerate at 0.20 m/s/s to complete it in 180 seconds.

Thanks.

 

[Redbelly98]

The runner accelerates at 0.20 m/s² for some time, then runs at constant speed for the remaining time.

Those two times must add up to ____?

Use the kinematic equations to figure out how far he travels during the accelerating portion, and also how far during the constant speed portion.

Those two distances must add up to _____?.

 

[leroyjenkens]

The runner accelerates at 0.20 m/s² for some time, then runs at constant speed for the remaining time.

Those two times must add up to ____?

Use the kinematic equations to figure out how far he travels during the accelerating portion, and also how far during the constant speed portion.

Those two distances must add up to _____?.

I only have one time and one distance to input into the equations. The time is 180 seconds and the distance is 1100 meters. If I put those into find his final velocity, it'll tell me the velocity he'll be at if he accelerates at .2 m/s/s for the entire 180 seconds. I'm not sure how to do it without knowing how long he accelerates. I know it's possible and I can see how I should have enough information to get the right answer, but the intuition for it just isn't coming to me.

Thanks for the response.

 

[Redbelly98]

I'm not sure how to do it without knowing how long he accelerates.

That is the tricky part. The key is to say he accelerates for t seconds, and proceed to write out the equations.

So, how far does the runner go if he accelerates at 0.2 m/s² for t seconds? (To answer this, you'll need the runner's initial velocity, at the moment he begins accelerating.)

 

[leroyjenkens]

That is the tricky part. The key is to say he accelerates for t seconds, and proceed to write out the equations.

So, how far does the runner go if he accelerates at 0.2 m/s² for t seconds? (To answer this, you'll need the runner's initial velocity, at the moment he begins accelerating.)

How do I incorporate him accelerating for t seconds into the other equations? Also, I seem to have two unknowns no matter what equation I put my known information in. I know his velocity when he first starts accelerating is 5.49 m/s. I know that he accelerates at .4 m/s/s. I know that the remaining distance is 1100 meters and the time he has left to make it to the finish is 180 seconds. But I can't figure out how to make use of that information, because it's not the distance or time he's accelerating, it's the total distance and total time left. Should I put my known information into one of the equations, get one variable alone on one side of the equation and then plug the entire other side of the equation into another equation in place of the variable?

 

[Redbelly98]

How do I incorporate him accelerating for t seconds into the other equations? Also, I seem to have two unknowns no matter what equation I put my known information in.

Yes, there are two unknowns: the time t and the distance d over which he accelerates. Since there are two unknowns, you will need two equations.

I know his velocity when he first starts accelerating is 5.49 m/s. I know that he accelerates at [STRIKE].4[/STRIKE] .2 m/s/s. I know that the remaining distance is 1100 meters and the time he has left to make it to the finish is 180 seconds. But I can't figure out how to make use of that information, because it's not the distance or time he's accelerating, it's the total distance and total time left.

Correct, he does not accelerate for the entire 180 seconds, and he does not travel the entire 1100 meters while accelerating. He accelerates for an unknown time t, and for an unknown distance d.

Should I put my known information into one of the equations, get one variable alone on one side of the equation and then plug the entire other side of the equation into another equation in place of the variable?

Yes. Can you write that first equation now? Just use "t" for the time, and use "d" for the distance. Which of the kinematic equations has distance and time, as well as the known quantities a=0.2m/s² and v_o=5.49m/s? Write that equation out.

 

[leroyjenkens]

Well this is the best one I've come up with and it still doesn't make sense. d = 5.49t + .1t^2.
To get that I had to use two different equations.
If I plug that into another equation I get 5.49t + .1t^2 = 5.49t + .1t^2 and that doesn't help me.
I've been doing physics problems in this book for a while. I had to think a little for each of them, but now all of a sudden I reach one that I'm mentally incapable of doing. The other ones took me a maximum of 20 minutes. I've been on this one for about a week now. Did I suddenly get brain damage? There's nothing more frustrating than this.

 

[Redbelly98]

I've been doing physics problems in this book for a while. I had to think a little for each of them, but now all of a sudden I reach one that I'm mentally incapable of doing. The other ones took me a maximum of 20 minutes. I've been on this one for about a week now. Did I suddenly get brain damage? There's nothing more frustrating than this.

This one is a more advanced difficulty level, so don't be too disheartened!

Well this is the best one I've come up with and it still doesn't make sense. d = 5.49t + .1t^2.
To get that I had to use two different equations.

That equation is correct. I'm surprised you had to use two different equations, in most physics classes one of the kinematics equations is

Δx = v_o t + (1/2) a t²​

and I am quite surprised if that is not among the equations your class is working with.

Be that as it may, the equation you wrote is correct and useful for proceeding with solving this...

If I plug that into another equation I get 5.49t + .1t^2 = 5.49t + .1t^2 and that doesn't help me.

True. The tricky part of this problem is that we should now consider the motion after the runner has finished accelerating. After accelerating, he runs at a constant speed for the remainder of the race.

So we need another equation that relates time and the distance traveled at constant speed, or a=0. As a hint, I will mention that this distance will be

1100 m - d​

and similar reasoning applies to the time it takes to run that distance.

 

[leroyjenkens]

This one is a more advanced difficulty level, so don't be too disheartened! That equation is correct. I'm surprised you had to use two different equations, in most physics classes one of the kinematics equations is

Δx = v_o t + (1/2) a t²​

Yes, I had that one but I kept ending up with a t term and a t squared term, so I tried the other ones but ended up with the same thing anyway.

True. The tricky part of this problem is that we should now consider the motion after the runner has finished accelerating. After accelerating, he runs at a constant speed for the remainder of the race.

So we need another equation that relates time and the distance traveled at constant speed, or a=0. As a hint, I will mention that this distance will be

1100 m - d​

and similar reasoning applies to the time it takes to run that distance.

I plugged the equation I got into the equation above, but I end up getting a quadratic no matter what I do. I used the quadratic formula to get 63.8. I'm not too good with quadratics so I'm not sure if that is equal to the T or not.
The problem with quadratics is there's so much room for error. I tried it again, doing it a different way in the same equation and got 200 instead of 63.8. I also got some negative answers, which I assume I can discard.
But I don't know if I'm supposed to even be doing a quadratic. So does this require me to do a quadratic and if so, does the answer I get equal the time? And if so, is that time equal to how long he accelerates?

 

[Redbelly98]

Yes, you will need to solve a quadratic equation to find t. That means you can get two solutions, but only one of them would be the actual answer.

Not sure why you're getting different answers though.

 

[leroyjenkens]

I solved the quadratic and got 63.48 seconds, but I can't figure out what that time tells me. It can't be how long he accelerates.

 

[Redbelly98]

Yes, the t you are solving for in the quadratic is how long he accelerates. That is exactly how we defined t for this problem (in Post #4, "...he accelerates for t seconds...")

Why do you think it means something else?

 

[leroyjenkens]

Yes, the t you are solving for in the quadratic is how long he accelerates. That is exactly how we defined t for this problem (in Post #4, "...he accelerates for t seconds...")

Why do you think it means something else?

I guess because it's such a high number. After I did it, I looked in the back of the book to find the right answer and it's 3.1 seconds. I got 63.48 seconds. I thought because it was such a huge difference, that I didn't actually find the duration of his acceleration. So I did get the wrong answer, right? If he really accelerated for that long, he'd be running like 40 miles per hour.

 

[notnimdab2009]

Hello , sorry to interrupt but i am searching for kinematics problems and came across this thread and i am interested also in solving and i want to share what i have done.

The runner has only 180 seconds left to complete the run with a distance of 1100 m. I used two formula here, 1 with acceleration and 1 with constant velocity. So for the first part of the solution, the runner must run say a distance d with an acceleration of 0.2 m/sec/sec for a time t. Then for the second part. The runner must run with constant velocity (his current rate which is 5.49 m/sec) for a distance of (1100 - d) meters and for a time of (180 - t) seconds.

1st part i used the formula : d = Vo (t) + .5at^2
Substituting the values : d = 5.49t + (.5)(0.2)t^2

2nd part i used the constant velocity formula d = v(t)
substituting the values : 1100 - d = 5.49 (180 - t)
I solved for d in this 2 nd part

and then equate the two equations d = d

i got t = 33.44 sec , this is the time needed for hte runner to accelerate at 0.2 m/sec/sec and then the remaining time of 180 - 33.44 sec = 146.56 sec he can then run at the constant rate of 5.49 sec.

I check my answers by getting the distance traveled and the two distances sum up to 1100 m .

I wasn't able to use the quadratic formula since 5.49t cancels out when i equate d = d.

Was my solution correct?

 

[Doubell]

what level is this type of question o or A level

 

[Redbelly98]

I guess because it's such a high number. After I did it, I looked in the back of the book to find the right answer and it's 3.1 seconds. I got 63.48 seconds. I thought because it was such a huge difference, that I didn't actually find the duration of his acceleration. So I did get the wrong answer, right? If he really accelerated for that long, he'd be running like 40 miles per hour.

Good point. Can you show your work? Otherwise I can't see where things went wrong.

2nd part i used the constant velocity formula d = v(t)
substituting the values : 1100 - d = 5.49 (180 - t)

5.49 m/s was the runner's initial speed at the beginning of the acceleration.

 

[leroyjenkens]

i got t = 33.44 sec , this is the time needed for hte runner to accelerate at 0.2 m/sec/sec and then the remaining time of 180 - 33.44 sec = 146.56 sec he can then run at the constant rate of 5.49 sec.

I check my answers by getting the distance traveled and the two distances sum up to 1100 m .

I wasn't able to use the quadratic formula since 5.49t cancels out when i equate d = d.

Was my solution correct?

Hi, thanks for your input. I don't think you got the correct answer, either. If he accelerated for 33.44 seconds, he would be running about 27 miles per hour. That seems a little too fast for a human.
I got 63.48 seconds, which is way too fast.
After I got that value, I got all excited that I finally answered this question I've been working on for a couple of weeks. I looked at the correct answer in the back of the book and it's 3.1 seconds.

Good point. Can you show your work? Otherwise I can't see where things went wrong.

Sure.
I got the value of 5.49t + .1t^2 as the initial distance, which I assume is the distance he runs during his acceleration period. Then I put that back into the same equation, but this time with a constant acceleration and a value of 1100 for the final distance.
That gave me a quadratic of .1t^2 + 10.98t - 1100 = 0. I put that into the quadratic formula and that gave me 63.48 seconds as the value of t.

 

[Redbelly98]

I got the value of 5.49t + .1t^2 as the initial distance, which I assume is the distance he runs during his acceleration period.

Yes, good.

Then I put that back into the same equation, but this time with a constant acceleration and a value of 1100 for the final distance.

I don't understand this statement. Are you now saying he runs with a constant acceleration for the full 1100 m? But you should be thinking in terms of a constant speed, for the remaining distance after he has finished accelerating.

That gave me a quadratic of .1t^2 + 10.98t - 1100 = 0. I put that into the quadratic formula and that gave me 63.48 seconds as the value of t.

It sounds like you haven't written a proper equation that describes the constant speed at the final portion of the run.

 

[leroyjenkens]

I don't understand this statement. Are you now saying he runs with a constant acceleration for the full 1100 m? But you should be thinking in terms of a constant speed, for the remaining distance after he has finished accelerating.

Sorry, I meant constant velocity. I think I was debating in my mind whether to type "constant velocity" or "zero acceleration" and combined them subconsciously.

Does that change anything, or is my equation still messed up?

 

[Redbelly98]

Sorry, I meant constant velocity.

Okay.

Does that change anything, or is my equation still messed up?

Since I had also gotten the book's 3.1s answer some days ago, it looks like your equation is incorrect. Perhaps you could show the steps of your work in between

d = 5.49t + .1t^2​

and

.1t^2 + 10.98t - 1100 = 0​

Of particular interest is the equation you wrote to express the constant speed part of the run.

 

[leroyjenkens]

The constant speed part is how I got the quadratic. I took the d = 5.49t + .1t^2 and put it back into the Δd equation, but this time I canceled out the second term in that equation, because a = 0 and instead of having d as an unknown, I made it 1100 to get .1t^2 + 10.98t - 1100 = 0.
At least I know the book isn't wrong. I've had that happen quite a few times before.

 

[Redbelly98]

Perhaps I wasn't clear. Please show (i.e., write it out) the equation that describes the constant speed part of the run. I really can't help with finding what's wrong if you don't do that.

 

[notnimdab2009]

I guess because it's such a high number. After I did it, I looked in the back of the book to find the right answer and it's 3.1 seconds. I got 63.48 seconds. I thought because it was such a huge difference, that I didn't actually find the duration of his acceleration. So I did get the wrong answer, right? If he really accelerated for that long, he'd be running like 40 miles per hour.

Hi, if the right answer is 3.1 sec. I tried to work this out backwards to check the "correct" answer.

Using the equation S=Vot +0.5at^2 to get the distance traveled with accleration for 3.1 secs,
s= (5.49)(3.1) + (0.5)(0.2)(3.1)^2 = 17.98 meters
And this would mean that the runner needs to run at his original constant velocity of 5.49 m/sec for the remaining 1082.02 meters (1100 - 17.98).

Now, i have the remaining distance(1082.02) and i have the constant velocity, 5.49 m/sec so it can give me the time needed to travel this distance and it should give me 176.9 seconds ( 180 - 3.1 ).

1082.02 = 5.49t
t= 197.09 which is greater than 176.9, therefore the runner will not make the run in less than 30 minutes. So i don't see why 3.1 seconds is the correct answer.

 

[Redbelly98]

... the runner needs to run at his original constant velocity of 5.49 m/sec for the remaining 1082.02 meters (1100 - 17.98).

No. Why do you think he would run at his original constant velocity? What does it mean, "to accelerate"?

 

[notnimdab2009]

oopss..now i realized...the problem perhaps as i interpret it now is that whatever the velocity he will achieved after the acceleration phase, this will be his velocity for the rest of the distance. I tried this but still i got only t=2.39 secs not 3.1 sec. I am really getting very interested in knowing the solution. Will continue to find the solution.

 

[notnimdab2009]

wow..i was so in a hurry to reply on my previous post that i missed some operations. Now i got the "correct" answer. It's 3.1328 secs. I am correct in the assumption that the velocity after the acceleration phase will be the velocity to be used for the rest of the distance. Thanks for this post. It's a very good problem.

 

[leroyjenkens]

Perhaps I wasn't clear. Please show (i.e., write it out) the equation that describes the constant speed part of the run. I really can't help with finding what's wrong if you don't do that.

Well, I found a mistake I made, so it's back to the drawing board.
Here's what I did, step by step.
I took the known values: 5.49 m/s and .2 m/s/s and plugged them into the Δd = Vot + .5at^2 equation.
I ignored initial distance since I made it 0 and that gave me d = 5.49t + .1t^2.
I put that back into the Δd = Vot + .5at^2 equation with the second term canceling out because a = 0. The mistake I made was making the initial velocity value 5.49. Since this part of the problem I'm trying to find his constant velocity after he accelerated, the initial velocity would be the velocity at the beginning of the constant velocity phase, which is a value I don't know.
So now I'm stuck with 1100 = 5.49t + .1t^2 + Vot and I don't know what to do. I'm yet again stuck with two variables.

 

[Redbelly98]

So now I'm stuck with 1100 = 5.49t + .1t^2 + Vot and I don't know what to do. I'm yet again stuck with two variables.

This is looking good, you are getting closer. Yes, that "Vo" term that is not 5.49 m/s is preventing you from finding t at this point.

Let's think about the acceleration part of the run again: he accelerates at 0.2 m/s² for t seconds. By how much would his speed have changed?

 

[leroyjenkens]

This is looking good, you are getting closer. Yes, that "Vo" term that is not 5.49 m/s is preventing you from finding t at this point.

Let's think about the acceleration part of the run again: he accelerates at 0.2 m/s² for t seconds. By how much would his speed have changed?

ΔV = at
So ΔV = .2t
Since ΔV is two values; initial velocity and final velocity, I took the initial velocity of 5.49 m/s and put it on the right side of the equation to get V = 5.49 + .2t. I used that value and the value of d, which is d = 5.49t + .1t^2 and put them both into the equation V^2 = Vo^2 + 2a(Δd).
I got pretty excited because it left me with just one variable; t. I put it all in, did all the multiplication and ended up with 30.14 + 2.196t + .04t^2 = 30.14 + 2.196t + .04t^2.
Apparently that doesn't work.
So to answer your question, during the acceleration part of the run, his velocity changes by .2t.
Or his velocity changes by (V + 5.49)/2.
I have all the equations, I seem to have tried everything, but nothing works. I keep ending up with X = X or a quadratic-looking equation that isn't a quadratic because it doesn't equal 0.

 

[Redbelly98]

The mistake I made was making the initial velocity value 5.49. Since this part of the problem I'm trying to find his constant velocity after he accelerated, the initial velocity would be the velocity at the beginning of the constant velocity phase, which is a value I don't know.

Right, we don't know the velocity at the beginning of the constant velocity phase. We do know the velocity at the beginning of the acceleration phase, 5.49 m/s, and...

ΔV = at
So ΔV = .2t

So, to sum that up, we know that the velocity was 5.49 m/s, and then it increased by an amount .2t. So the velocity during the constant velocity phase would be ____?

 

[leroyjenkens]

So, to sum that up, we know that the velocity was 5.49 m/s, and then it increased by an amount .2t. So the velocity during the constant velocity phase would be ____?

The velocity was 5.49 m/s at the beginning, so I'll make that my initial velocity and that would make the velocity during the constant velocity phase... v = 5.49 + .2t

 

[Redbelly98]

Yes. So at that velocity, what distance does the runner go during the constant velocity phase?

 

[leroyjenkens]

Yes. So at that velocity, what distance does the runner go during the constant velocity phase?

The only formula I can plug that into and it make sense is d = (vo + v)/2 times t, and that gives me something pretty crazy...

d = Vot/2 + 2.745t + .1t^2

 

[Redbelly98]

The only formula I can plug that into and it make sense is d = (vo + v)/2 times t, and that gives me something pretty crazy...

d = Vot/2 + 2.745t + .1t^2

No, remember at that point he is going at a constant velocity of 5.49+0.1t. (And t is the time during which he accelerated earlier.)

There is a pretty simple formula that gives distance when traveling at constant velocity. Again, the key here is it is constant velocity.

 

[Redbelly98]

Another hint: distance = velocity x 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 ...

.
.
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.