1. ## average speed

A jet pilot plans to cover a 1000mi course at an average speed of 1000mph. For the first 800 mi the speed is 800mph. At what rate must the remaining distance be covered. no ans given

well I did it this way(since it is an average)

$\frac{\frac{800}{1} + \frac{200}{h}}{2}=\frac{1000}{1}$

$800+\frac{200}{h} =2000$

$800h + 200 = 2000h$

$200 = 2000h-800h$

$\frac{200}{1200} = h$

$h = \frac{1}{6}$ or $1200$mph for the rest of the $200m$

just seems there would be a shorter way

2. Hello, bigwave!

This is a trick question . . .

A jet pilot plans to cover a 1000-mile course at an average speed of 1000 mph.
For the first 800 miles, the speed is 800 mph.
At what rate must the remaining distance be covered?

He plans to fly 1000 miles at an average of 1000 miles per hour.
. . So his flight will take exactly one hour ... right?

He flies the first 800 miles at 800 miles per hour.
. . This takes one hour.

This leaves no time for him to fly the other 200 miles.

[Maybe he can say, "Beam me over, Scotty" ?]

3. ## yes but

yes but

if the pilot is averaging the speed of 1000mph that doesn't mean that he has to complete the trip in one hour does it... he can go slower than the average and then go faster than average to go the 1000 miles.. the 800miles in one hour got me going to..
I just didn't think time was fixed here just the average....

or am I just walking in the woods...

r

4. hi there

what soroban said was correct, if he plans to finish the trip in one hour, then why does the question say he is going at 800mph. thats one hour gone for you already...

5. where does it actually say that he needs to finish the trip in one hour??? average speed does not determine the time spend he could go over an hour and still have an average speed of 1000mph. In order to finish the last 200 miles he needs to go 1200mph and that means over an hour.. it is asking to maintain an average not a set rate. just being cranky...

6. It says 1000mi at 1000mph which means the trip takes 1 hr.

$\displaystyle 1000mph=\frac{1000miles}{1hr}\rightarrow \ \mbox{therefore, in 1 hour the pilot goes 1000miles}$

7. Originally Posted by bigwave
A jet pilot plans to cover a 1000mi course at an average speed of 1000mph. For the first 800 mi the speed is 800mph. At what rate must the remaining distance be covered. no ans given

well I did it this way(since it is an average)
Here is your basic problem- there are many different kinds of "average" and you chose the "arithmetic average" of the two speeds. That is not correct. An "average speed" is "distance divided by time". There is NO "dividing by 2" in that.

$\frac{\frac{800}{1} + \frac{200}{h}}{2}=\frac{1000}{1}$

$800+\frac{200}{h} =2000$

$800h + 200 = 2000h$

$200 = 2000h-800h$

$\frac{200}{1200} = h$

$h = \frac{1}{6}$ or $1200$mph for the rest of the $200m$

just seems there would be a shorter way

8. Originally Posted by bigwave
A jet pilot plans to cover a 1000mi course at an average speed of 1000mph. For the first 800 mi the speed is 800mph. At what rate must the remaining distance be covered.
You'd be ok if: the first 800 mi was covered in 59 minutes and 59 seconds.
Then you'd beed to cover the last 200 mi in 1 second !!

9. ok thanks everyone....

10. Hence, his only option is to go faster
on the return journey and cheat by taking an average
over twice the distance.

11. You've lost me here. Are you thinking about including the motion of the earth turning from west to east? What reason do you have to think that the speeds given are not "ground speeds", relative to the earth? And even if they are "air speeds", relative to the air, you would only need to take into account the local wind speed.

12. well one thing I didn't quite get over this discussion (altho have to admit entertaining)

was how can any jet fly 800 miles at 800mph and with that no hope ever of seeing an average speed of 1000mph even if he ignites the boosters..

doesn't the pilot have to fly more than that to get the average speed up there (or suddenly fly backwards or sideways as mentioned) ????

or is this jet just suddenly dropping out of the sky and Parachuting (etc a dud word problem)

I am losing sleep over this.... help (I confess my division by 2 was sinful)

13. The initial questions implies he covers the 1000 miles in 1 hr. If you go 800miles at 800mph (1hr), you can't cover the other 200miles and have it take no time. At least in real world applications.

14. Originally Posted by bigwave
well one thing I didn't quite get over this discussion (altho have to admit entertaining)

was how can any jet fly 800 miles at 800mph and with that no hope ever of seeing an average speed of 1000mph even if he ignites the boosters..

doesn't the pilot have to fly more than that to get the average speed up there (or suddenly fly backwards or sideways as mentioned) ????

or is this jet just suddenly dropping out of the sky and Parachuting (etc a dud word problem)

I am losing sleep over this.... help (I confess my division by 2 was sinful)
If he has to go an exact route, 1000 miles in distance, then he's run out of options,
because if he covers the exact 1000 miles, then he now has to cover it in more than 1 hour
(on account of using up a full hour to cover the first 800 miles).
If his average speed over the first 800 miles is 800mph, then he's taken an hour to cover the distance.
To have flown a distance of 1000 miles at an average speed of 1000mph,
he would have taken an hour for the entire flight.

Hence, the only way to cover the route at an average speed of 1000mph is to cover more than 1000 miles,
either by taking a detour (to increase the distance and fly at a higher speed) or by including a return journey.
However, given the wording of the question (a 1000 mile course), it's cheating.

The figures given in the question require the entire course to be covered in 1 hour
to give an average speed of 1000mph for a 1000 mile course.
The average is "entire distance divided by entire time taken".
The average speed for the first 800 miles is "800 divided by time taken".
We can only use a time > 1 hour if we increase the distance of the course also.

15. Some teacher probably made that problem up as a joke...