# Homework Help: His average speed for the first 4km is what?

1. Apr 11, 2014

### chikis

A man runs a distance of 9km at a constant speed for the first 4km and then 2km/h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4km is what?

Total distance = 9km
Total time = 1 h

For the 1st for 4km, he moved at a constant speed.

For the last 5km, his speed was increased by 2km/h.

Let the costant speed at 4km be xkm/h.

At the last 5km, his speed became (x + 2) km/h.

The total speed for the whole runs took him
[(x+2) + x] km/h
= (2x + 2) km/h

speed s = distance d/time t

s = d/t

t = 9/(2x + 2)

1 = 9/(2x + 2)

2x + 2 = 9

2x = 7

x = 7/2 = 3.5 km/h

Therefore the speed for the 4 km = 3.5 km/h

Since the speed was increased by 2km/h, that means the speed for the last 5 km/h would become (3.5 + 2)
= 5.5 km/h.

But when I checked, my answer did not seem right. Please I need help. Thank you.

2. Apr 11, 2014

### jbriggs444

This seems to assume that the first half and the last half of the run took equal amounts of time. But that assumption may not be true. [An "average" speed would normally be weighted based on time -- for instance, 2 km/h for 20 minutes and 4 km/h for 40 minutes would average to 3 1/3 km/h]

How much time did the first 4 km take in terms of x?
How much time did the last 5 km take in terms of x?
What do these two times add up to?

Solve for x.

3. Apr 11, 2014

### nasu

Assuming equal time is not the only problem.
Adding the speeds for the two segments does not make any sense whatsoever.
Or for any segments.
Why not add the speeds for any minute of that hour? You will get a much larger speed.

You need to write the equations for each segment, as suggested by jb.

4. Apr 11, 2014

### HallsofIvy

If he ran 4 km at speed x km/hr, then he took 4/x hrs.

If he ran 5 km at speed x+ 2 km/hr, then he took 5/(x+ 2) hrs.

Altogether he ran 9 km in (4/x)+ 5/(x+ 2) hours but we are told he took one hour.

Solve (4/x)+5/(x+ 2)= 1.

5. Apr 12, 2014

### chikis

4/x +(5/x 2) = 1

9x+8 = x2+2x

x2 - 7x - 8 = 0

(x 1)(x-8) = 0

x = 8 or x = -1

The correct answer is 8 km/h.

My observation here is that we can only make equation for the total time and use it to find the speed. Why is that if we make equation for the total distance (as I did in my previous work), we cannot use it to get the accurate value for x the speed?

6. Apr 12, 2014

### Staff: Mentor

Sure you can, but you have to set your equations up correctly.

For the first part of his run, we have 4 = r * t
For the second part of his run, where he runs faster, we have 5 = (r + 2)(1 - t)

In these equations, r is his running speed in the first leg, and t is his time for the first leg. On the second leg of his run, his speed is r + 2 (in km/hr), and his time is 1 - t (in hours).

If you solve the first equation for t, and substitute it into the second equation, you eventually end up with the same quadratic equation you have.

BTW, using x for the speed is not a very good choice. It's better to use a letter that makes it more obvious what it represents, which is why I chose r (for rate) and t (for time).

7. Apr 13, 2014

### vela

Staff Emeritus
Your approach was flawed. As nasu pointed out, adding the two speeds doesn't make any sense.

Consider these examples: Suppose you drive at 10 km/h for two hours and then 20 km/h for another hour. The total distance you drove would be 40 km, and the total time would be three hours. Your average speed would be 40/3 km/h, about 13.3 km/h. This result should seem reasonable. The average speed is between the lowest speed and the highest speed. It tends to be closer to the lower speed because you drove longer at that speed than at the higher speed.

Suppose instead you drive at 10 km/h for one hour and then 20 km/h for two hours. This time, the total distance is 50 km, and your average speed would be 50/3 km/h, about 16.7 km/h. This result also seems reasonable. Again, the average speed is between two speeds. It tends to be closer to the higher speed this time because you drove longer at that speed than at the lower speed.

Now based on your reasoning, the "total speed" is 30 km/h in both cases. Does this make sense? No. For one thing, in both cases, the car always went slower than that speed. Second, you have the same total speed in both cases. Unlike the average speed and the total distance travelled, the so-called total speed apparently doesn't depend on how much time was spent at either speed. So what exactly is this speed supposed to represent physically? The answer is it doesn't represent anything physically meaningful.

8. Apr 13, 2014

### chikis

Please expanciate a little bit. What is the thinking behind the expression (1 - t) ?

Last edited: Apr 13, 2014
9. Apr 13, 2014

### SammyS

Staff Emeritus
If t represents the time to complete the first 4 km and the total time taken to complete the whole 9 km is 1 hour, how long does it take to run the last 5 km ?

10. Apr 13, 2014

### chikis

I know that t represent time. Am now asking
1-t is what?

11. Apr 13, 2014

### vela

Staff Emeritus
You should be able to reason that out yourself based on SammyS's hint.

12. Apr 13, 2014

### Staff: Mentor

No, t represents the amount of time to run the first 4Km.
From my earlier post...
Also see Sammy's hint in post #9.

13. Apr 14, 2014

### chikis

(1-t) is nothing more than subtracting the time for the first 4km from the total time, which gives the time for the last 5km.

14. Apr 15, 2014

### SammyS

Staff Emeritus
Yes !

So, now you can get an expression (in terms of t) for the speed over the first 4 km and another expression (in terms of t) for the speed over the remaining 5 km . You also know how these two expressions are related to each other, which allows you to get an equation with both these expressions in it. This equation has only one variable, t, so solve for t.

15. Apr 17, 2014

### chikis

Solving that way we make us to have two unknowns and that will make the problem too dificult to solve.

16. Apr 17, 2014

### Staff: Mentor

You didn't read what I wrote in post 6.
After making the substitution, you have a quadratic equation with one variable.