MHB I just want to know one divergent formula

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    Divergent Formula
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Train A travels at 60 mph and Train B at 75 mph, making B 5/4 times faster than A. Since both trains leave simultaneously, Train B will never reach a point where it has traveled twice the distance of Train A. The relationship between their speeds indicates that B will always cover a lesser distance ratio compared to A over time. The formula for distance, d = st, confirms that for B to be twice as far as A, its speed would need to be double that of A, which is not the case. Therefore, it is impossible for Train B to travel twice as far as Train A under the given conditions.
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Hello. I asked my professor and he couldn't figure it out. If train A and B leave the same point at the same time, A traveling 60mph, B traveling 75mph, how long will it take for B to have traveled twice as far as A?
 
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justone said:
Hello. I asked my professor and he couldn't figure it out. If train A and B leave the same point at the same time, A traveling 60mph, B traveling 75mph, how long will it take for B to have traveled twice as far as A?

Train B is traveling 5/4 as fast as train A. If they depart at the same time, then train B will always have traveled 5/4 as far as train B...there is no point in time (where $0<t$) for which train B will have traveled twice as far as train A.

I am going to move this thread to our algebra forum, as that is a better fit.
 
Recall that $d=st$, where $d$ is distance, $s$ is speed and $t$ is time.

With the speeds of train A and train B denoted as $s_1$ and $s_2$ respectively, we must have

$s_2t=2s_1t\implies s_2=2s_1$, but this is clearly not true.
 
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