and returns at the same speed. At the same time another man leaves from Point A to point B at an average speed of 30 miles an hour and returns at an average of10 miles an hour. Who gets back first? Commonsense tells me that they will get back at the same time, both having averaged 20mph but a collegue insists that one would get back earlier and there is a discrepancy I can't work out. The second person will be half-way back when the first person arrives at point B but then will reduce his speed to a third, having a quarter of the distance left. Can anyone explain this in a simple mathematical model and put us out of our misery?
2007-01-27 02:37:47 · 16 answers · asked by checkmate 6 in Science & Mathematics ➔ Mathematics
I answered a similar question before. They would arrive together if the person (in the UK we are not allowed to say man without adding "or woman") doing 30MPH changed to 10MPH at the half way point in time not the half way point in distance.
2007-01-27 04:56:03 · answer #1 · answered by David P 4 · 0⤊ 2⤋
Suppose the distance from Point A to Point B is 60 miles. It would take the person traveling 20 mph 3 hours to go to Point B and 3 hours to come back, a total of 6 hours. The second person would need 2 hours to get to Point B, but 6 hours to get back, for a total of 8 hours. Plug in any number for the distance between A and B and you'll find that the first person takes less time.
Average speed is not the average of the speeds traveled. (i.e. it's not (speed 1 + speed 2)/2.) It's the distance traveled divided by the time taken to travel (average speed = d/t).
2007-01-27 02:50:16 · answer #2 · answered by star_gazer10 2 · 1⤊ 0⤋
Take the distance as 30 miles
lHe travels 20mph for 60 miles takes 3 Hours
Second man drives 30 miles taking 1 Hour
He drives back at 10mph taking 3 Hours (Total) 4Hrs
You must include distance in the equation
2007-01-27 03:48:12 · answer #3 · answered by Taffy 1 · 0⤊ 0⤋
Average speed is not calculated just by adding outgoing and incoming speed.It depends on the distance travelled and hence time taken.
if the distance is 1 mile ,the first man takes 1/20 +1/20 or 1/10 hrs to travel 2 miles.Hence his average speed is 2 divided by 1/10 which is 20 miles per hour
the second man takes 1/30 hrs to go 1 mile and 1/10 hrs to come back.So total time taken by him is 1/30 +1/10=4/30 hrs
Therefore his average sped is 2 divided by 4/30=15Miles per hour.
As the average speed of the first man is bettr,he would come back earlier
2007-01-27 03:01:49 · answer #4 · answered by alpha 7 · 0⤊ 0⤋
Rate of travel is calculated by the formula distance/time, whereas you are trying to divide rate by distance to arrive at a figure for the average rate of travel. Stick to the formula rate = distance/time, and you'll be fine.
For example, let's say point A and B are 20 miles apart. The A man, traveling 20 mph, finishes the trip in two hours. The B man finishes the first part of the trip in 40 minutes, and the second part of the trip takes 2 hours. 2.67 hrs is significantly different from 2 hrs. While the first man travels at a consistent rate of 20 miles / hour, the second man has traveled 40 miles in 2.67 hours, for an average speed of 14.98 miles / hour.
There's your discrepancy.
2007-01-27 02:47:32 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Let AB = d miles
First Man: Total distance travelled by him = 2d miles at an average speed of 20 miles / hour.
The time t(1) taken by him is: t(1) =2d /20 =d /10 hours.
Second Man: He travels the distance of d miles , going from A to B, at an average speed of 30 miles per hour and, again, he travels the distance of d miles , going from B to A, at an average speed of 10 miles per hour . The time t(2) taken by him is:
t(2) =d /30 + d /10 = 2d /15 hours.
So t(2) > t(1)
2007-01-27 02:53:44 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Lets say point B is 20 miles from point A:
Driver A (20mph) - A --> B = 20miles @ 20 mph = 1 hour
B ---> A = 20miles @ 20mph = 1 hour = 2hrs total time
Driver B - A--->B = 20 miles @ 30mph = 2/3hr or 40 min
B--->A = 20 miles @ 10mph = 2 hours = 2hr 40 min total time
Driver A @ constant 20mph gets back first.
Hope that helps.
2007-01-27 02:52:03 · answer #7 · answered by Na Pomoč 3 · 0⤊ 0⤋
I have no simple mathematical model, but I just tried out an example in which the distance was 20 miles there and back. They one who stays at 20 mph gets ther about 40 minutes before the other. Just use example and keep plugging in numbers
A-B=20miles, B-A=20miles
A-B, person 1, at 20mph - 1hr B-A,person 1, at 20mph- 1hr
A-B, person 2,at 30mph- 40minutes B-A,person 2, at 10 mph- 2hr
person A- 2hrs person B- 2hrs 40 minutes
a 40 minute difference
2007-01-27 02:49:48 · answer #8 · answered by Soccer Midget 2 · 0⤊ 0⤋
Set the distance as 60 miles. Man A (traveling at 20 mph) will complete the round trip in 6 hours(3 hours there and 3 hours back) whereas man B (travelling at 30 mph there and 10 mph ont he return) will take 8 hours to complete the round trip (2 hours there and 6 hours back)
2007-01-27 02:46:53 · answer #9 · answered by CHRIS P 3 · 0⤊ 0⤋
Put in simple terms:
Assume the distance from point A to point B is 30 miles
20 mph from A to B will take 1.5 hours
20 mph from B to A will take 1.5 hours
Traveling at 20 mph on both the first and second leg will have a round trip time of 3 hours
30 mph from A to B will take 1 hours
10 mph from B to A will take 3 hours
Traveling at 30 mph on the first and 10 mph on the second leg will have a round trip time of 4 hours
2007-01-27 02:54:39 · answer #10 · answered by Anonymous · 1⤊ 0⤋
If point A to B is 20 miles.
First man takes 2 hours to travel 40 miles.
2nd man travels 30 miles in 1 hour. Therefore it only takes him 40mins to get to B. So on the return journey he has 20 miles to cover at 10 mph which will take 2 hrs. a total of 2hrs 40mins.
If the distance was 30 miles between points.
1st man's return journey take 3 hrs.
2nd man takes 4hours.
Speed change occurs at the turnaround point not after time lapse according to your information.
Are you daft?
2007-01-27 02:50:27 · answer #11 · answered by laughingspam 3 · 1⤊ 0⤋