Distance travelled during acceleration.

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Discussion Overview

The discussion revolves around a physics homework problem involving a cyclist's motion, specifically focusing on the distance traveled during acceleration, the work done, and the concept of constant velocity. Participants explore calculations related to these aspects while clarifying the problem statement.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant calculates the distance traveled during acceleration using the formula s = 1/2(u+v)t and arrives at an initial result of 72 m, later correcting it to 18 m.
  • Another participant questions the calculation steps and seeks clarification on how the distance was derived.
  • Work done during acceleration is calculated as W = FxD, leading to a value of 648 joules based on the displacement and force.
  • Participants discuss the concept of constant velocity, with one suggesting it is derived from the tractive resistance force, calculated as 16 N.
  • There is confusion regarding the problem statement, with participants noting that the question appears incomplete and questioning the need to find the constant velocity when it is given as 3 m/s.
  • One participant asserts that the constant velocity is indeed 3 m/s and suggests that work done during this period is calculated using the frictional force and displacement.

Areas of Agreement / Disagreement

Participants express uncertainty about the completeness of the problem statement and the necessity of finding the constant velocity. There is no consensus on the interpretation of the question or the calculations presented.

Contextual Notes

Participants highlight potential limitations in the problem statement, noting that it seems to lack explicit questions and may lead to confusion in determining the required calculations.

The Futur
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Homework Statement



A cyclist accelerates from rest to a velocity of 3 m/s in 12 seconds, continues at this speed for one minute and then freewheels to stop. If the total mass of cyclist and cycle is 80 kg, and the tractive resistance rate may be assumed constant at 0.2 N/kg.,

Homework Equations


distance.
s= 1/2(u+v)t


The Attempt at a Solution



my attempt was

s= 1/2(0+3)12
s=1/2(3)12
s=6x12
s= 72

im not too sure if that's right but i tryed my best :)
 
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The Futur said:
s= 1/2(0+3)12
s=1/2(3)12
s=6x12
s= 72
How did you get from s=1/2(3)12 to s=6x12?
 
Fightfish said:
How did you get from s=1/2(3)12 to s=6x12?

my bad s= 1/2(u+v)t
s= 1/2(0+3)12
s= 1/2(3)12
s= 1.5(12)
s= 18m

Work done during acceleration and constant velocity

W=FxD
w= 36x18
w= 648 joules

now how to find the constant velocity?

is it

c.v => (Mass x Tractive resistance rate)
c.v= 80x0.2
c.v= 16N ?
 
Hm...what exactly are you trying to calculate from the scenario (what's the question asking for?)
16N is the tractive resistance force (ie. friction). Where does this constant velocity thingy come from?
 
Fightfish said:
Hm...what exactly are you trying to calculate from the scenario (what's the question asking for?)
16N is the tractive resistance force (ie. friction). Where does this constant velocity thingy come from?

second part of the question.
 
Is it just me or is the question spliced off after "...assumed constant at 0.2 N/kg.,"? Cos' from the the spliced extract, I only see a scenario and no questions.
 
Fightfish said:
Hm...what exactly are you trying to calculate from the scenario (what's the question asking for?)
16N is the tractive resistance force (ie. friction). Where does this constant velocity thingy come from?

Is the Distance traveled during acceleration correct?

The constant velocity come from the second part of question (iii).

Question:
(iii)Determine The work done during acceleration and the constant velocity.

Work done during acceleration

W=FxD
w= 36x18
w= 648 joules

now how to find the constant velocity?

my try was.

c.v => (Mass x Tractive resistance rate)
c.v= 80x0.2
c.v= 16N ?
 
The constant velocity is, as given in the question, 3m/s.
Work done during this period is then simply the frictional force X displacement since forward driving force = frictional force (magnitude)
 
if the constant velocity is, as given 3m/s. why do they ask me to find the constant velocity?
 

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