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Question

A bullet train leaves Kyoto for Tokyo traveling 240 miles per hr. at noo.

Ten minuts later another train leaves Tokyo for Kyoto traveling 150 miles per hr.

If Tokyo & Kyoto are 300 miles apart, at what time will the trains pass each other?

Answer 1

Distance = (Rate)(Time)

 

When the two trains pass each other, the sum of their distances is

300 miles.

 

Let x = time (in hours) traveled by first train since noon

Then, x - 1/6 = time (in hours) traveled by second train since noon                             (note that 10 minutes = 1/6 hr)

 

Distance of first train since noon = 240x

Distance of second train since noon = 150(x - 1/6) = 150x - 25

 

So, 240x + 150x - 25 = 300

 

                      390x = 325

 

                          x = 5/6 hr = 50 minutes 

 

The trains will pass at 12:50 PM.  

 

 

Answer 2

Hi.  Hope this helps.

 

Let's define our problem:

Total distance (d) = 300 miles (from Tokyo to Kyoto)

S₁ = Speed of Train 1 (240 mph)

S₂ = Speed of Train 2 (150 mph)

T₁ = Departure time of Train 1 = 0 hours (equals zero since this is the baseline time)

T₂ = Departure time of Train 2 = 1/6 hours (delay time relative to train 1)

 

C = Collision Time (the time the trains will pass each other).

 

If the departure time of train 2 then we need to know how many miles are covered by train 1 before its departure.

Train 1 has a rate of travel of 240 mph.  After 10 minutes (1/6 hours), train 1 has traveled 1/6(s₁)

Lets make this equal to x.

x = d - 1/6(S1) = 260 miles

 

This helps us understand the x distance between both trains when they are both in motion.

 

Now we need to define the velocity (v) of travel.  Since the trains are traveling in opposite directions, the velocity would be the sum of the two trains speeds.

Therefore  v = S1 + S2 = 240 + 150 = 390 mph.

This combined speed tells us how much of our x distance is being reduced each hour.

 

Our last component is time (t).  This is what we want to find out.

Therefore:

x = v * t

 

This means the distance traveled = velocity times time.

We know the distance traveled and the velocity so let reconfigure the equation in terms of time.

t = x/v

 

Now we substitute our x and v equivalents factored earlier:

t = (d - 1/6(S₁)) / (S₁ + S₂)

 

To solve we substitute our known values for d and S and solve

t = (300 - 1/6(240)) / (240 + 150)

t = 260/390 = 2/3 hours past noon

 

C = 12 + t

C = 12:40 pm (collision time)