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Sagot :
We have three different velocities. Downhill velocity (Dv), Level ground velocity (LGv), and Uphill velocity (Uv). We need to find the level ground velocity to find the time. Initially, the question says that:
[tex]\text{Dv = 120 miles per hour}[/tex]Then we have "A train goes twice as fast downhill as it can go uphill". Mathematically this means:
[tex]2\times Uv\text{ = Dv}[/tex]We can find Uv:
[tex]Uv=\frac{120}{2}=60\text{ miles per hour}[/tex]The next information is "and 2/3 feet as fast uphill as it can go on level ground". 2/3 in decimal is 0.6666, so we can round it to 0.67.
2/3 feet as fast means that Uphill velocity is greater than level ground velocity. We can say that:
[tex]1.67\text{ }\times\text{ LGv = Uv}[/tex]We can find LGv:
[tex]\text{LGv = }\frac{60}{1.67}=36\text{ miles per hour}[/tex]Now, to find the time we can apply the medium velocity equation above:
[tex]\text{Velocity = }\frac{Dis\tan ce}{\text{Time}}[/tex][tex]36\text{ = }\frac{45}{\text{Time}}[/tex][tex]\text{Time = }\frac{45}{36}[/tex][tex]\text{Time = 1.25 hour or 75 minutes}[/tex]
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