How long does it take it increase speed?

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To determine how long it takes for a freight train with a mass of 1.2e+7 kg to accelerate from rest to 82 km/h under a constant pull of 8.4e+5 N, the relevant formula is t = (mass * velocity) / force. After converting 82 km/h to m/s, the calculation yields a time of approximately 5.1e-2 minutes. However, some participants question whether this duration is realistic for such a large train to reach that speed. The discussion emphasizes the importance of correctly applying the physics formulas and unit conversions. Overall, the calculations and assumptions surrounding the time required for acceleration are critically analyzed.
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Homework Statement


A freight train has a mass of 1.2e+7 kg. If the locomotive can exert a constant pull of 8.4e+5 N, how long does it take to increase the speed of the train from rest to 82 km/h?


Homework Equations


(mass*velocity)/time=force


The Attempt at a Solution


(1.2e+7 * 82)/h = 8.4e+5

i solved for h
h = (8.4e+5)/(9.84e+8)
h = 8.54e-4

then i had to convert from hours to min.
8.54e-4*60

so i got
5.1e-2 min

this is my first time posting on this forum so i hope i followed the right format.
 
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82km/h = \frac{82*1000m}{3600s}
 
can you explain the above?
 
I just converted the speed from km/h to m/s which is usually what you should do.

82km/h

in 1h train travels 82km
=> in 1h train travels 82*1000m
in 3600s train travels 82000m
in 1s train travels 82000/3600 m

therefore the velocity in m/s is 820/36
 
that doesn't answer my question, i think I am messing up on the formula. I have one submission left so i don't want to risk it
 
F=\frac{mv}{t} \Rightarrow t=\frac{mv}{F}

t=\frac{(1.2*10^7 kg)*((\frac{82000}{3600} ms^-1)}{8.4*10^5 N}



Can you calculate now?


But your answer was 5.1*10^{-2}minutes. Doesn't that seem like a small amount of time to go from 0 to 82km/h ?
 
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