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)   (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
Multiply by (x)(x+10 km/hr)

Subtract 720xkm from each side.

-x^2-10x+7200=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=28900 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: -90, 80. Here's your graph:

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
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