Distance travelled during acceleration.

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A cyclist accelerates from rest to 3 m/s in 12 seconds, then travels at that speed for one minute before stopping. The distance covered during acceleration is calculated using the formula s = 1/2(u+v)t, resulting in 18 meters. The work done during this acceleration phase is determined to be 648 joules, based on the force and distance. The constant velocity is confirmed to be 3 m/s, as stated in the problem. The discussion highlights confusion regarding the question's clarity and the relevance of calculating the constant velocity.
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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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