A train stops for equal amounts of time at 4 stations bewtween A and B which are 40 km apart. THe traveling time between these two points is 20 minutes. At most how long can the train spend at each station if it is to average at least 60km/hr for the entire trip including stops.
how do you do this problem?
2007-01-06 16:10:40 · 8 answers · asked by lauta 2 in Science & Mathematics ➔ Mathematics
Let time for which it stops at one station be x min
Net time it stops = 4x min
Average speed = 60km/hr = 1km/min
Distance to cover = 40 km
Distance/average speed = total travelling time
Net travelling time = (40 + 20 - 4x) min
Total travelling time = 40 min
40 = 60 - 4x
4x = 20
x = 5 min
The problem is too ideal ...acceleration should be humungous it it were true!
2007-01-06 16:24:46 · answer #1 · answered by Som™ 6 · 0⤊ 0⤋
The average is 60 km/hr, or 1 km/min - and the traveling time from A to B is 20 min, the distance from A to B is 20 km. How can the distance between stations be 40 km. If the distance was 40 km between stations, the total distance is 120 km. He would have to go 120 km/hr without stopping. Something is off here.
2007-01-06 16:29:41 · answer #2 · answered by Anonymous · 0⤊ 0⤋
To average 60 km/hr, it would need to go a distance of 40 km in 40 minutes.
60 = 40 / t
t =2/3 hour = 40 minutes
So if the train needs 20 minutes to get from point A to point B, then it can spend a total of 40-20 = 20 minutes at rest.
Splitting this up evenly among 4 stations = 5 minutes.
2007-01-06 16:25:48 · answer #3 · answered by z_o_r_r_o 6 · 0⤊ 0⤋
This is pretty simple if you consider it in terms of overall time. You need to average 60 km/hr or 1 km/minute over 40 km so you have 40 minutes to get there. Travel time is 20 minutes so that gives 20 minutes of stopping time at the four stations for 5 minutes stop at each station.
2007-01-06 16:38:01 · answer #4 · answered by Pretzels 5 · 0⤊ 0⤋
Well, just thinking it through...
If the train went straight through without stopping, it would average:
40 km/20 min x (60min/hr) = 120 km/hr
That's twice as fast as it needs to go.
So the train can afford to spend as much time stopped as it spends travelling, i.e. 20 minutes.
Since it can spend 20 minutes stopped, divided by 4 stations, it can spend 5 minutes at each station.
Double check:
20 minutes in transit + 4 stops x 5 minutes per stop = 40 minutes in transit.
40 km/40 min x (60 min/hr) = 60 km/hr....check.
(of course, to be fair, this assumes that the train can go from 0 km/hr to 120 km/hr instantaneously, which isn't realistic, but it's just a math problem.....)
2007-01-06 16:22:08 · answer #5 · answered by Jim Burnell 6 · 0⤊ 0⤋
If the train goes 40 kilometers in 20 minutes, it is going 120 km.p.h. Now, you must find out how long the train can go for 120 km.p.h. and 0 km.p.h. in 20 minutes, and average to 60. Let X be time. (120x+0X)/2=20 120x=40. x=1/3. You can stop at each station for 20 seconds, and your total average speed will be 60 km.p.h.
2007-01-06 16:23:58 · answer #6 · answered by Anonymous · 0⤊ 0⤋
d = r*t
t = d/r, so the elapsed time to average 60 kph is
t = (40 km)(60 min/hr)/(60 km/hr) = 40 min
let T = max time at each stop.
T = (40 - 20)/4
T = 20/4
T = 5 minutes
2007-01-06 16:25:01 · answer #7 · answered by Helmut 7 · 0⤊ 0⤋
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2016-11-27 01:06:52 · answer #8 · answered by Anonymous · 0⤊ 0⤋