How long does it take train B to catch up to train A?

Question

Train A leaves a station traveling at 80 kilometers per hour. Eight hours later, train B leaves the same station traveling in the same directionat 100 kilometers per hour.

Answer 1

There are four ways to account for the lead Train A has. I'm trying not to complicate things but to show there are multiple ways to get to the answer; all of which follow a common procedure. In the end I'll show a trick that's much quicker but the work below will help clarify what is happening and why different people can all show different methods but still get the correct answer. The four methods are...

 

1)Train A starts 8 hours ahead

2)Train B starts 8 hours behind

 

(An 8 hour lead puts Train A 640km ahead {80km/h*8h}. )

3)Train A travels 640km less

4)Train B travels 640km farther

 

The important thing to note is that each train has its own d=v*t equation (can also be a t=d/v equation seen in cases 3/4).

d_a=v_a*t_a

d_b=v_b*t_b

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You can set the distances equal to each other, but you must then relate the times.

1)t_a=t_b+8   :Train A is 8 hours ahead (or Train A leaves at t_b=-8)

2)t_b=t_a-8    :Train B is 8 hours behind (or Train B leaves at t_a=8)

In 1 and 2 we're using time as a variable (that is relating the times) we must solve for d or distance in the equation. In 3 and 4 we use distance as a variable (using the relative distances) and we'll solve the equation for time after this

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d_a=d_b

v_a*t_a=v_b*t_b

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1)v_a*(t_b+8)=v_b*t_b

80*(t_b+8)=100*t_b

640=(100-80)t_b

640=20t_b

t_b=32  (where t_b is the amount of time Train B travels before it catches Train A)

 

2) v_a*t_a=v_b*(t_a-8)

80*t_a=100*(t_a-8)

80*t_a=100*t_a-800

800=(100-80)t_a

800=20t_a

t_a=40 (where t_a is the amount of time Train A travels before it is caught, notice its 8 hours longer because its been traveling longer)

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Now that we solved for the time in either case 1) or case 2) we have the answer, and if we wanted we could easily find the distance each train traveled before they met.

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t_a=d_a/v_a

t_b=d_b/v_b

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Just like we set the distances equal we can also set the times equal and then we must relate the distances.

3) d_a=d_b-640       :Train A travels 640km less to meet at the same point  (as it has a 640km head start)

4) d_b=d_a+640       :Train B travels 640km farther to meet at the same point

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t_a=t_b
d_a/v_a=d_b/v_b
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3)(d_b-640)/80=d_b/100

100*(d_b-640)=80*d_b

(100-80)d_b=640*100

20d_b=64,000

d_b=3,200 (where d_b is the distance Train B travels to catch Train A)

 

4)d_a/80=(d_a+640)/100

100d_a=80(d_a+640)

(100-80)d_a=80*640

20d_a=51,200

d_a=2,560 (where d_a is the distance Train A travels while Train B is catching up, notice it is smaller than d_b because Train A already traveled 640km before Train B and 3200-2560=640)

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Now we've solved for the distance traveled but the question asks for time traveled so we divide by speed to get time t=d/v.

3200/100=32

2560/80=32   :and we get 32 hours for both trains

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Below are graphs for each case

1) (t_b , d_b) When Train B (blue line) leaves the station (y=0) Train A has already been traveling for 8 hours and is at y=640 (where y is the distance travelled)

http://www.wolframalpha.com/input/?i=100t%3D%2880%29%28t%2B8%29

2) (t_a, d_a) Train A (blue line) leaves the station (y=0) at time=0 (x=0), and Train B (red line) leaves at time=8 (x=8)

http://www.wolframalpha.com/input/?i=80t%3D%28100%29%28t-8%29

3) (d_b , t_b) When Train B (Red Line) leaves the station (x=0) Train A (Blue line) has already been travelling for 8 hours (y=8)

http://www.wolframalpha.com/input/?i=%28d%2B640%29%2F100%3Dd%2F80

4) (db, tb) When Train B (Red line) leaves the station (x=0) Train A (blue line) is already 640km ahead (at x=640).

http://www.wolframalpha.com/input/?i=%28d-640%29%2F80%3Dd%2F100

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If you've made it this far hopefully you've learned a thing or two. Now for the easy part.

t=d/v is just the velocity equation rewritten

Δt=(Δd)/(Δv) is the same equation but written using Δ or the difference in. The advantage here is all three terms here are easy to find and will be the solution

Δt=difference in time from Train B leaving the station to meeting Train A : The solution

Δd=difference in position at the beginning : 8hours*80km/h=640km

Δv=difference in train speed : 100-80=20km/h

and...

Δt=640/20=32 hours. AKA the answer

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What is happening here is that we are using a moving reference frame. By using a reference frame that moves with Train B, were only concerned with their relative motion/position which is that Train A starts 640km away and Train B is 20km/h faster. This is the same as if Train B was stationary and Train A starts moves toward Train B at 20km/h from 640km away. So all we need to do is find out how long it takes something moving 20km/h to travel 640km.

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so remember

Δt=(Δd)/(Δv)

Answer 2

What the trains have in common is the time it takes them to meet, which is for both the same time.

When train B leaves, train A is already gone for 8 hours and has covered a distance of 8h x 80km/h, which is 640 km.

 

After that it takes them the same amount of time (t) to meet.

 

640km + t x 80km/h = t x 100km/h

 

Solve for t; first you cancel it out on the right side by dividing both sides by t:

640km/ t  + 80km/h = 100km/hr

 

subtract 80km/h from both sides.

640km/t = 100km/h - 80km/h

640km/t  = 20km/h

 

Now solve for t:

 

640km/20km/h = t

t = 32h

Answer 3

Train B leaves station after 8 hours.

By that time train A travels= 80*8=640km

Say, after that train A travels distance x km, such that it meets train B

So time taken by train A = x/80hrs

Time taken by train B =(x+640)/100

So,

(x+640)/100=x/80

 

x=2560km

 

Time when they meet

2560/80=32 hours

 

So they meet after 32 hours