How long does it take it increase speed?

  • Thread starter Thread starter RanHam335
  • Start date Start date
  • Tags Tags
    increase Speed
Click For Summary

Homework Help Overview

The problem involves calculating the time required for a freight train to accelerate from rest to a speed of 82 km/h, given its mass and the force exerted by the locomotive. The subject area pertains to dynamics and kinematics in physics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the relationship between force, mass, and velocity to find the time. Some participants provide conversions for speed from km/h to m/s and question the validity of the original poster's calculations.

Discussion Status

Participants are exploring different aspects of the problem, including unit conversions and the appropriateness of the formula used. There is a concern raised about the reasonableness of the calculated time, indicating a productive direction in the discussion.

Contextual Notes

The original poster expresses concern about making a mistake with their submission, highlighting the pressure of limited attempts in the homework context.

RanHam335
Messages
3
Reaction score
0

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.
 
Physics news on Phys.org
82km/h = [itex]\frac{82*1000m}{3600s}[/itex]
 
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
 
[tex]F=\frac{mv}{t} \Rightarrow t=\frac{mv}{F}[/tex]

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



Can you calculate now?


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

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K