Answer :
Here is the solution:
Start with the fact that both trains have made the same distance= 's'.
In physics distance is a multiplication of speed of your train = 'v' and period of time in which it was moving= 'T' (this is only true when the speed doesn't change during movement, or we talk about avarage speed like in your task).
Then:
[tex]s= v_{D} * T_{D}[/tex]
"D" symbol refers do diesel train
Freight train moves with 15[km/h] greater speed and it travels 1 [h] shorter then:
[tex]s = v_{F} * T_{F} = (v_{D}+15)*(T_{D}-1)[/tex]
As the distance made by both train is the same you can compare the right parts of both equations:
[tex]v_{D} * T_{D}=v_{F} * T_{F} = (v_{D}+15)*(T_{D}-1)[/tex]
Diesel train moved 6 hours and freight train 5 hours then:
[tex]v_{D} * 6=(v_{D}+15)*5[/tex]
[tex]6*v_{D}=5*v_{D}+75[/tex]
[tex]6*v_{D}-5*v_{D}=75[/tex]
[tex]v_{D}=75[/tex]
Here is your answer: The average diesel train speed is 75 km/h
Start with the fact that both trains have made the same distance= 's'.
In physics distance is a multiplication of speed of your train = 'v' and period of time in which it was moving= 'T' (this is only true when the speed doesn't change during movement, or we talk about avarage speed like in your task).
Then:
[tex]s= v_{D} * T_{D}[/tex]
"D" symbol refers do diesel train
Freight train moves with 15[km/h] greater speed and it travels 1 [h] shorter then:
[tex]s = v_{F} * T_{F} = (v_{D}+15)*(T_{D}-1)[/tex]
As the distance made by both train is the same you can compare the right parts of both equations:
[tex]v_{D} * T_{D}=v_{F} * T_{F} = (v_{D}+15)*(T_{D}-1)[/tex]
Diesel train moved 6 hours and freight train 5 hours then:
[tex]v_{D} * 6=(v_{D}+15)*5[/tex]
[tex]6*v_{D}=5*v_{D}+75[/tex]
[tex]6*v_{D}-5*v_{D}=75[/tex]
[tex]v_{D}=75[/tex]
Here is your answer: The average diesel train speed is 75 km/h
