a bird can travel 90 miles and hour an train travels 60 miles and hour

Question

a bird can travel 90 miles and hour an train travels 60 miles and hour if the bird and train leave at the same time 90 miles apart and the bird fly to the train and goes back the the station how do we figure out how many miles the bird fly if it goes back and forth to the train and the station until the train gets to the station

Answer 1

This is a D-I-R-T (Distance-Is-Rate-times-Time, or D=RT) Problem.  With such problems, two variables are equal or related so that we can solve the problem.  In this problem, the bird and the train travel until they meet (the sum of their travel distances equals the initial separation) or else the bird travels back to the Station.

It usually helps to draw a picture:   [note: we will re-use x often]

 

1) The bird and the train travel toward each other

Station

Bird->                                      90 miles                                                         <- Train

|---------|---------|---------|---------|---------|---------|---------|---------|---------|

 

The time the bird flies and the time the train runs are equal, so let’s set them equal:

                               Bird                          Train

 T = D/R    (x miles)/(90 mph)   =     (90-x)/60       [Let x=bird miles]

            60x = 90(90-x)         [cross-multiply]

            60x = 8100 – 90x           [distribute]

          150x = 8100                 [add 90x to both sides]

               x = 54 miles        [to meet: the bird flew 54 miles, train went 36 miles]

          The train is now also 54 miles from the Station.

 

2) The bird flies back to the station while the train continues toward the Station.

Station               54 miles                  <-Bird

                                                       <- Train

|---------|---------|---------|---------|----|

                               Bird                        Train

 T = D/R   (54 miles)/(90 mph)    =   (x)/60      [Let x=new train miles]

        90x = 60(54)                  [cross-multiply]

        90x = 3240                

           x = 36 miles

   The train has traveled 36 miles closer to the station, so is now (54-36)=18 miles away.

 

3) The bird and the train travel toward each other.

Station

Bird->  18 mi <- Train

|---------|-------|

                               Bird                       Train

 T = D/R   (x miles)/(90 mph)  =  (18-x)/60    [Let x=new bird miles]

       60x = 90(18-x)        [cross-multiply]

       60x = 1620 – 90x         [distribute]

     150x = 1620         [add 90x to both sides]

           x=10.8 miles

  The train is now also 10.8 miles from the Station.

 

4) The bird flies back to the station while the train continues toward the Station.

Station         <-Bird

      10.8 mi  <- Train

|---------|-------|

                               Bird                        Train

 T = D/R   (10.8 miles)/(90 mph)    =   (x)/60      [Let x=new train miles]

     90x = 60(10.8)                  [cross-multiply]

     90x = 648                      [add 90x to both sides]

        x = 7.2 miles

   The train has traveled 7.2 miles closer to the station, so is now (10.8-7.2)=3.6 miles away.

 

5) The bird and the train travel toward each other.   [get the idea, yet?]

 

Let’s create an Excel spreadsheet calculation:

Step      Bird miles       Bird position    Train miles   Train position
START       0                     0                   0                   90
1             54                   54                  36                  54
2             54                    0                   36                  18
. . .

15        0.0006912      0.0006912      0.0004608      0.0006912
TOTAL  134.994816

Now, 0.0006912  is 3.7 foot.  This bird is really quick, but at 60 mph the train travels at 88 ft/sec, so measuring distances gets somewhat ridiculous (this is asymptotic).  The bird effectively flies 135 miles before the train smashes it into the Station. 

Answer 2

the bird is at the station

the bird travels at 90 mph for as long as the train is traveling

once the train gets to the station and stops, the bird stops flying

the train travels at 60 mph for 90 miles

d=rt

90=60t

t=90/60

t=1.5 hours

the bird flies for 1.5 hours because the train gets to the station in 1.5 hours and when the train gets to the station the bird stops flying

d=rt

d=90*1.5

d=135 miles is the distance the bird flies

 

 

Answer 3

Okay I think your saying that the train and the bird starts off from the opposite ends and the initial distance between them is 90 miles.

Let's keep this formula in mind

d = r*t 

since they are traveling from the opposite ends and heading towards each other, we know that the time it takes for them to meet is quite small 

90= 60t + 90t

90 = 150t

t = 90/150 hour (this tells us the time it takes for them to meet)

if we plug in the time for the formula (for bird) we can find out how much distance the bird traveled within that time period

d = 90*(90/150) = 54 miles 

meaning the bird meets the train when the bird traveled about 54 miles whereas the train (90-54 = 36 miles) or plug in the same time (90/150) into the formula for train

d = 60*(90/150) = 36 miles

The bird once hits train has to head back to the station meaning it will travel the same distance twice; therefore, 54*2 = 108 miles. (the answer)

We can assume that the bird will have enough time to travel back and forth the full 108 miles before the train hits the station since the time it has for the train to hit full 90 miles is greater than the time the bird takes to travel 108 miles.