A person walks up a hill at a speed of 8.0m/s. Once he reaches the top, he walks back down (same distance) at a speed of 12.0m/s. What is the average speed for the trip? Give answer and reason why/calculations. Thanks!
2006-09-07 14:24:19 · 7 answers · asked by zillian.thrax 2 in Science & Mathematics ➔ Physics
t = d/8 + d/12 = 20d/96
D = 2d
Vave = D/t = 2d/(20d/96) = 9.6 m/s
2006-09-07 14:33:50 · answer #1 · answered by Steve 7 · 3⤊ 0⤋
Let's do it this way.
Suppose the distance that the man walks to reach the top of the hill is 2400m. He takes 300 secs (2400/8) to go up. Then he walks down the same distance and the time he takes is 200 secs(2400/12). So the total time to go up and come down a distance of 2400+2400=4800 m is 300+200=500 secs
The average speed is 4800 m / 500 sec = pl work it out if you need to..
2006-09-07 21:39:28 · answer #2 · answered by rabi k 2 · 1⤊ 0⤋
Assume the distance up the hill is x.
Speed = x/t = 8 (going up) so it took t-up = x/8 seconds to go up.
Going down it took t-down = x/12 seconds.
Total time = t-up + t-down = (x/8 + x/12) = 20x/96 seconds
Total distance = 2x
Average speed = total distance/total time = 2x/(20x/96) = 9.6 m/s
2006-09-07 21:33:54 · answer #3 · answered by ic3d2 4 · 2⤊ 0⤋
(speed) = (distance)/(time elapsed)
(time elapsed) = (distance)/(speed)
Let's call the total overcome distance '2x' = x (one way)+ x (way back).
The time spent to reach the top of the hill is x/8. The time to come back to the starting point is x/12. So the total elapsed time to go back and forth is (x/8+ x/12).
(avg. speed for the trip) = (total distance)/(total elapsed time) =
___= (2x)/(x/8+ x/12)
___= 2x/(3x/24+ 2x/24) <--- basically algebra from now on
___= 2x/(5x/24)
___= 2x*24/5x
___= 48/5
___= 9.6 m/s
2006-09-07 21:41:50 · answer #4 · answered by Illusional Self 6 · 1⤊ 0⤋
Say the distance is "[d] (meters)"
Time = distance / rate
The time for the uphill trip is :
T1 = [d] (meters) / [8.0] (meters /second) = [d / 8.0] (seconds)
Time for down hill is:
T2 = [d] (meters) / [12.0] (meters / seconds) = [d / 12.0] (seconds)
Average speed is found by using the quantities for the ENTIRE trip.
Distance = [2d] (meters)
Time = T1 + T2 = [ [d/12.0] + [d/8.0] ] (seconds)
= [ [8d + 12d] / [8*12] ] (seconds) = 20d / 96
= [5d/24] (seconds)
Again, Rate = Distance / time.
Average speed = [2d] (meters) / [5d/24] (seconds)
=[24*2d / 5d] (meters / second)
=[48/5] (meters/second)
=9.6 m/s
2006-09-07 21:57:09 · answer #5 · answered by Anonymous · 0⤊ 0⤋
i dont even take physics
and i know the answer
10 m/s is the average
why??
take the average of the 2 numbers by adding then dividing by 2 because there are 2 numbers that were added
2006-09-07 21:26:18 · answer #6 · answered by shorty 3 · 0⤊ 4⤋
EF this, mathematical physics is math... true physics lies in the theoretical realm
2006-09-07 21:30:21 · answer #7 · answered by Kevin M 3 · 0⤊ 3⤋