# How Fast Does the Racecar Driver Have to Drive?

1. Dec 15, 2013

### Medgirl314

1. The problem statement, all variables and given/known data

A racecar driver must average 200 km/h over the course of a time trial lasting ten laps. If the first nine laps were done at 199 km/h, what average speed should be maintained for the last lap?

2. Relevant equations

avg =distance/time

3. The attempt at a solution

He drove at 199 for 9 laps.

199*9=1791

He needed to average 200 for 10 laps.

2000

2000-1791=209.

The answer is 209.5 km/h, but why?? I also tried dividing 199 by 9 and 200 by 10, averaging to get 21.05, and then multiplying by 10, but that isn't working either. I'm having a hard time finding an appropriate equation, so I was trying to think it through logically. It didn't work.

2. Dec 15, 2013

### Curious3141

Simple averaging doesn't work because the average speed cannot be calculated (in general) by a simple arithmetic mean between two or more speeds. The other things you tried make even less sense.

The average speed of a journey is always total distance travelled/total time taken.

Let the speed on the final lap be v. Let a single lap be a a distance of d. What you need to do is find expressions for the total distance and the total time and divide them. You will find the d cancelling out in the final expression. Equate this to the required average speed and solve.

3. Dec 15, 2013

### Medgirl314

Hi Curious3141,

I'm not sure I'm understanding your directions. I know dividing comes in, and both the final lap and each individual lap comes in. D will cancel out so that I acquire an average speed. I'm not really sure what you mean by "What you need to do is find expressions for the total distance and total time and divide them." The main thing I am having a hard time figuring out is how I can get the time so I can divide the total distance by the time,which I neglected to state in my post.I think that was my original equation, but I'm not sure how to find the time. If you could walk me through that part a bit more, that would be great! Just a disclaimer, this isn't actually an assignment for school. It's just an extra problem I picked out of my textbook to practice, and I can't seem to find an explanation for it. Thanks again!

4. Dec 15, 2013

### Curious3141

distance for one lap = "d" km. So distance for 9 laps (in terms of d) = ?

time taken = distance/speed. He drove 9 laps at 119 km/h. So time taken for those 9 laps (in terms of d) = ?

He drives the last lap at "v" km/hr. Time taken for this last lap = ?

Total time taken = ?

Total distance = ?

Average speed over all 10 laps (in terms of d and v) = ?

"d" should cancel out here.

Now equate that to 200 km/h and do the algebra.

Last edited: Dec 15, 2013
5. Dec 16, 2013

### Medgirl314

distance for one lap = "d" km. So distance for 9 laps (in terms of d) = d=9km

time taken = distance/speed. He drove 9 laps at 119 km/h. So time taken for those 9 laps (in terms of d) = ? d=0.76

He drives the last lap at "v" km/hr. Time taken for this last lap = ?

Total time taken = ?

Total distance = ? 10 km

Average speed over all 10 laps (in terms of d and v) = ?

"d" should cancel out here.

Now equate that to 200 km/h and do the algebra.

I put in what I know so far, but I'm not sure how to find the time here: He drives the last lap at "v" km/hr. Time taken for this last lap = ?

Thanks!

6. Dec 16, 2013

### nasu

The distance for 9 laps is 9d and not 9 km.
This is what Curious meant by "express in terms of d". d is not known but you don't need it.
And the time for these 9 laps is 9d/(119 km/h). Again, in terms of d.

Now express the time for the last lap and so on.

Last edited: Dec 16, 2013
7. Dec 20, 2013

### Medgirl314

Sorry about the wait! I'll have to work on this this weekend as I have assignments for physics and this is extra. I don't expect an answer this weekend though, as it is nearly Christmas. So Merry Christmas! (-:

8. Dec 28, 2013

### Medgirl314

distance for one lap = "d" km. So distance for 9 laps (in terms of d) = d=9d
time taken = distance/speed. He drove 9 laps at 119 km/h. So time taken for those 9 laps (in terms of d) = ? d=0.76

He drives the last lap at "v" km/hr. Time taken for this last lap = ?

Total time taken = ?

Total distance = ? 10d

Average speed over all 10 laps (in terms of d and v) = ?

"d" should cancel out here.

Now equate that to 200 km/h and do the algebra.

That's what you meant by "In terms of d", correct? I thought I could just choose whatever unit I wanted to make it simple and then leave it out of my answer.

9. Dec 28, 2013

### 1MileCrash

d is not equal to 9d. Do not write "d=9d," because that is false. If you write false statements in your work, you are going to make mistakes.

The distance for one lap is d. Thus, the distance for 9 laps is 9d.

No. You wrote a value for d, which is the distance of a lap. We are looking for time, and this time will be in terms of d.

We know that the distance for 9 laps is 9d. He drove 9 laps at 119 km/h. Thus, he drove a distance of 9d at 119 mph. You have distance and speed, so what is the time?

Do not try to find a value for d, this is both useless and impossible. You are expressing other things using the term d.

Last edited: Dec 28, 2013
10. Dec 28, 2013

### Medgirl314

It was a typo. :-)

I think there is a slight miscommunication here. All I'm really looking for is for someone to help me figure out the algorithm that will lead to the answer I gave. I can piece together what I need to know, but I don't know how to get there. If someone could help me come up with the steps, I would be forever grateful. Thanks!

11. Dec 28, 2013

### 1MileCrash

We generally don't provide solutions, but will help walk you through. The outline given by Curious will bring you to the answer rather quickly.

He said:
"time taken = distance/speed. He drove 9 laps at 119 [sic] km/h. So time taken for those 9 laps (in terms of d) = ?"

As he said, time taken = distance/speed. He drove 9 laps at 199 km/h. Time is just distance/speed, so divide your expression for distance (9d) by your speed (199), which is 9d/199. That's it.

There is no reason for you to try and find a numeric value for d, the racecar driver can do this regardless of how long each lap is - it doesn't matter. We are asking you to express things algebraically so that you can build an equation to solve - we aren't asking for numbers.

Now, he says:
He drives the last lap at "v" km/hr. Time taken for this last lap = ?

Just like I did above, can you express the time taken for the last lap, knowing that time = distance/speed? Again, do not try to find numeric values for anything yet.

12. Dec 28, 2013

### Medgirl314

Oh, sorry, by algorithm I just meant the steps. Okay, that makes more sense. I thought we were trying to find numbers already.

Do you mean something like d/200 ? Since the last lap was driven at 200 km/h?

13. Dec 28, 2013

### 1MileCrash

No.

The last lap was driven at a speed of "v." We do not know v's value yet, that is what we solve for at the end.

Since the last lap was driven at a speed of "v" and the distance of a lap is "d", what is the time for this final lap (again, no numbers in this part)?

14. Dec 28, 2013

### Medgirl314

Ah, I forgot that part.

d/v

15. Dec 28, 2013

### 1MileCrash

Yes!

So, d/v is how long the final lap took.
And, 9d/199 is how long the first nine laps took.

So how much time did the entire race take (still in terms of d and v)?
And how long was the entire race (in terms of d)?

Since we know the length of the entire race, and the time the entire race took, what is an expression for the average speed of the whole race?

16. Dec 28, 2013

### Medgirl314

Time for entire race: (9d/199)+(d/v)

Length for entire race: 10d

Average speed: (10d/[9d/199]+[d/v])

17. Dec 28, 2013

### 1MileCrash

Yep

(though, your average speed is missing brackets - but I think you know how its written)

Now, we have that:

Average speed = 10d / ([9d/199]+[d/v])

Now, can you see that the d's in the numerator and denominator cancel now? Convince yourself of that, and then we have the following:

Average speed = 10 / ([9/199]+[1/v])

Now, we want the average speed to be equal to 200. So let the average speed be 200, and solve for v.

18. Jan 4, 2014

### Medgirl314

Does that mean equating "v" to 200?

Thanks again!

19. Jan 4, 2014

### haruspex

No, it means equating average speed for the 10 laps to 200.

20. Jan 8, 2014

### 1MileCrash

Edit: whoops, didn't see that haruspex already responded to that.

Sorry for the late response

No; you are solving for v. You equate average speed to 200. The problem is asking "what is v whenever average speed is 200?"

When someone says "solve for v" in an equation, it means put v on one side of the equation by itself. Set average speed to 200, and get v on one side of the equation by itself. That will be the result, because v is what we are trying to find.

You have to be more confident - tell the equation what you want. Remember that v stands for our speed during the final part of the race, that's what we want to find out