A man travels to work at a speed of 50 mph and returns home at 30 mph. What was the average speed during the round trip?
2007-01-31 04:05:51 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
We do not know how many miles it takes it to go to work and back
2007-01-31 04:15:23 · update #1
First, the answer is not 40mph. Because the man travels longer at 30 mph than he does at 50 mph. Average speed is not a simple average, but a weighted average, in which component velocities are weighted by their duration. (The normal intuition is to weight the component velocities by the distance traveled, but the math doesn't work that way. Consider this: A man takes a ski lift up a mountain at 30 mph. How fast must he ski back down to attain an average speed of 60 mph?)
So, the original problem. We don't know how far he travels, but it helps to think about it. Let's make each leg of the trip take D miles; it'll cancel out when we finish.
So, the morning commute takes D/50 hours and the evening commute takes D/30 hours. The total time is D/50 + D/30 hours. Collect this into one fraction:
D/50 + D/30 = (D/50)*(3/3) + (D/30)*(5/5) = 8D/150 = 4D/75
So, the total time is 4D/75. Average speed is then (Total distance)/(Total time), so that's (2D) / (4D/75) = 2*75/4 = 37.5 mph.
2007-01-31 04:37:05 · answer #1 · answered by happymutant42 1 · 1⤊ 0⤋
37.5 mph
Assume ANY distance to work. Say, 50 miles (for round figures). The trip in takes exactly 1 hr. Traveling at 30 mph on the way home, 50 miles will take 1.66666 hours. In the day he spent 2.66666 hours traveling 100 miles. 100 miles / 2.66666 hours = 37.5 mph. Again, any distance will work.
2007-01-31 04:21:37 · answer #2 · answered by Brett B 2 · 1⤊ 0⤋
Since we don't know the distance of the trip, let's call it d. Speed = distance / time, so time = distance/speed.
The travel going to work took time t1 = d/50, and the travel back took t2 = d/30, for a total of 2d distance in (t1 + t2) time, so the average speed is
v = 2d / (t1 + t2) =
2d / (d/50 + d/30) =
2d / ((30d + 50d) / 1500) =
2d * 1500 / 80d =
3000/80 =
300/8 =
37.5
2007-01-31 07:56:50 · answer #3 · answered by jcastro 6 · 1⤊ 0⤋
Let time for outward journey = t1 hours and time for return journey be t2 hours
Let distance to work be D miles
t1 = D/50 and t2 = D/30
Time for whole journey = t1 + t2 = D/50 + D/30
= 3D/150 + 5D/150 = 8D/150 = 4D/75
Average speed = distance/ time = 2D /4D/75 = 2D x 75/4D = 1/2 x 75 = 37.5 mph
2007-01-31 04:22:30 · answer #4 · answered by Como 7 · 1⤊ 0⤋
12-15 isn't too quickly...on a highway bike its to no longer quickly...on a intense mountain bike...its now and returned difficult My precise speed on a flat highway is 35 yet in basic terms for some seconds on my highway bike, yet my accepted speed is 17 while i glance at my cyclometer. I journey long distances and my accepted is going down with the aid of the years... so counting on how lots adventure you have...your holiday might flow swifter. additionally in the journey that your ridding around the city automobiles can get interior the way alongside with pedestrians and stoplights so perhaps discover a distinctive direction than you may take using.
2016-11-01 23:23:03 · answer #5 · answered by ? 4 · 0⤊ 0⤋
average speed=(30+50)/2=40mph
2007-01-31 04:20:10 · answer #6 · answered by Math gal 2 · 0⤊ 1⤋
(50 plus 30) divide 2, so 40 mph.
Since it's the same distance covered each way you don't need to know it, it's just a simple average out of the two numbers.
2007-01-31 04:10:40 · answer #7 · answered by rchlbsxy2 5 · 1⤊ 1⤋
U are not supposed to average an average u might be OK here sense both samples are supposedly equal.
2007-01-31 04:18:47 · answer #8 · answered by JOHNNIE B 7 · 0⤊ 0⤋
(2*50*30)/(50+30) = 37.5
2007-01-31 04:10:02 · answer #9 · answered by Anonymous · 0⤊ 1⤋
40mph?
2007-01-31 04:11:47 · answer #10 · answered by Anonymous · 0⤊ 1⤋