SOLUTION: a train covers a distance of 720 km at a speed of x km/h. On the return journey, it travels at an average speed of y km/h and completes the whole journey in an hour less. if the sp

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Question 957496: a train covers a distance of 720 km at a speed of x km/h. On the return journey, it travels at an average speed of y km/h and completes the whole journey in an hour less. if the speed during the return journey was 10 km/h more, what were the speeds in m/s?
is there a formula to use?

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
a train covers a distance of 720 km at a speed of x km/h. On the return journey, it travels at an average speed of y km/h and completes the whole journey in an hour less. if the speed during the return journey was 10 km/h more, what were the speeds in m/s?
distance=720km; original speed=x km/hr; return speed=y km/hr=x+10 km/hr
%28720km%2Fx%29-1hr=720km%2F%28x%2B10%28km%2Fhr%29%29 Multiply by (x)(x+10 km/hr)
%28720km%29%28x%2B10%28km%2Fhr%29%29-%281hr%29%28x%29%28x%2B10%28km%2Fhr%29%29=720xkm
720xkm%2B7200km%5E2%2Fhr-%28x%5E2hr%2B10xkm%29=720xkm Subtract 720xkm from each side.
7200km%5E2%2Fhr-x%5E2hr-10xkm=0
-x^2-10x+7200=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -1x%5E2%2B-10x%2B7200+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-10%29%5E2-4%2A-1%2A7200=28900.

Discriminant d=28900 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--10%2B-sqrt%28+28900+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-10%29%2Bsqrt%28+28900+%29%29%2F2%5C-1+=+-90
x%5B2%5D+=+%28-%28-10%29-sqrt%28+28900+%29%29%2F2%5C-1+=+80

Quadratic expression -1x%5E2%2B-10x%2B7200 can be factored:
-1x%5E2%2B-10x%2B7200+=+-1%28x--90%29%2A%28x-80%29
Again, the answer is: -90, 80. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-1%2Ax%5E2%2B-10%2Ax%2B7200+%29

x=80 Original speed was 80 km/hr.
80km/hr+10km/hr=90km/hr Return speed was 90 km/hr.
80km/hr(1000m/km)(1hr/3600sec)=22.22 m/sec Original speed was 22.22 meters per second.
90km(1000m/1km)(1hr/3600sec)=25 m/sec Return speed was 25 meters per second.
CHECK:
(720km/80km/hr)-1hr=720km/90km/hr
9hrs-1hrs=8hrs
8hrs=8hrs