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A train leaves a town traveling at a speed of 16 mph.

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3 Answers

The question you were given is poorly worded.
 
From the standpoint of the FIRST train, it takes 40 minutes before it is caught.
 
It only takes the SECOND train, 40-30 = 10 minutes to catch the FIRST train.

I agree that the way the question is worded that it appears to be asking for the 10 minutes, not the 40, but I will show you how to get the "40 minute" answer.
 
 
 
 
The distance for each train must equal the same for the second train to have "caught up".
 
Let t be the time in hours that the first train will travel until it is caught.
 
 
D1 = (16mph) (t)
 
 
The second train will travel for .5 LESS than t hours or t - .5
 
 
D2 = (64mph)(t - .5)
 
 
 
D1 = D2
 
16t = 64t - 32
 
Add 32 to both:
 
16t + 32 = 64t
 
Subtract 16t from both:
 
32 = 64t - 16t
 
32 = 48t
 
Divide both by 48:
 
32/48 = t
 
t = 2/3  of an hour or 40 minutes.
Let us denote time the second train will travel as X
then the first train will travel X + 0.5 hour.

The distance = speed times time.

Trains catch up when they will travel the same distance.

The First train distance D1= 16mph (X + 0.5)h O.5 hours = 30 minut.
The second train distance D2 = 64mph * X h.

16(X + 0.5) = 64X

16X + 8 = 64X

16X + 8 - 16X = 64X - 16X

8 = 48X X = 1/6 hour or 10 minutes. Answer: 10 min
 

D1 = 16mph * (30min + 10 min) = 16mph * 0.66 hour = 10.6 miles
D2= 64mph * 1/6h = 10.6 miles

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From the point of view of the second train, the first catches it up with the speed s=64-16=48 mph. When the second train starts, the first train is 16*0.5=8 miles away. Thus it will take the second train 8/48=1/6 hour or 10 minutes to catch the first train.

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Are you sure you stated the problem correctly? Because with the given numbers none of your answers is correct.

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