What is the speed of the bicycle?

  • #1
Homework Statement:
Find the speed of the bicycle using the given data to up to 2 significant figures.
Relevant Equations:
I am not understanding how to proceed. I could use a hint for calculating the distance travelled by the car when the car is decelerating and then calculating the distance and speed of the bike will be elementary and can be solved easily. Thanks(or sorry for a possible mistake in phrasing or formatting my question) in advance.
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Answers and Replies

  • #2
PeroK
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How far did you get with this? It looks a bit tricky.
 
  • #3
How far did you get with this? It looks a bit tricky.
Use basic integration, specifically the polynomial rule and the fundamental theorem of calculus.
 
  • #4
PeroK
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Use basic integration, specifically the polynomial rule and the fundamental theorem of calculus.
Where did you get stuck?
 
  • #5
jbriggs444
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You seem to have pasted in an image containing your work rather than having typed in your work directly. Let me transcribe it for you.
On integrating f(a) in the first phase where the v is constant (since the acceleration is equal to 0 m/s2, I get ##v_{t_1}## is equal to 12 m/s and ##v_{t_2}## is equal to 0 m/s (given). Also you get ##x_{t_1}## is equal 12 m after integrating the velocity function.
I agree that ##v_{t_1} = v_{t_2} = 12 m/s##.
I agree that ##x_{t_1}## is equal to 12 meters. This is 12 m/s multiplied by 1 s. That integration is trivial.

Given that starting point, what prevents you from writing down a formula for ##v(t)## that is valid for ##t_1 \le t \le t_2##?

Worst case, what prevents you from drawing the velocity/time graph on graph paper and counting squares?
 
  • #6
Like I said, I got stuck in the phase of motion during which the car undergoes deceleration. You will be able to understand better if you read the questions.
 
  • #7
PeroK
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Like I said, I got stuck in the phase of motion during which the car undergoes deceleration. You will be able to understand better if you read the questions.
I have read the question. Where did you get stuck, precisely?
 
  • #8
You seem to have pasted in an image containing your work rather than having typed in your work directly. Let me transcribe it for you.

I agree that ##v_{t_1} = v_{t_2} = 12 m/s##.
I agree that ##x_{t_1}## is equal to 12 meters. This is 12 m/s multiplied by 1 s. That integration is trivial.

Given that starting point, what prevents you from writing down a formula for ##v(t)## that is valid for ##t_1 \le t \le t_2##?
You have not used LaTeX properly so it is hard to understand.
 
  • #9
jbriggs444
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You have not used LaTeX properly so it is hard to understand.
It is proper. I fixed one closing ## and it is all perfect.
 
  • #12
jbriggs444
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You should use "$$", not ##
Embedded ##\LaTeX## works just fine, thank you. With the forum software here, $$ starts the LaTeX on a separate line while ## displays in-line.

But yes, refresh is sometimes necessary.
 
  • #13
It was not coming properly before, so I refreshed the page.
 
  • #14
You seem to have pasted in an image containing your work rather than having typed in your work directly. Let me transcribe it for you.

I agree that ##v_{t_1} = v_{t_2} = 12 m/s##.
I agree that ##x_{t_1}## is equal to 12 meters. This is 12 m/s multiplied by 1 s. That integration is trivial.

Given that starting point, what prevents you from writing down a formula for ##v(t)## that is valid for ##t_1 \le t \le t_2##?

Worst case, what prevents you from drawing the velocity/time graph on graph paper and counting squares?
I am supposed to use integration. I am unable to use an equation of motion using any three variables. I have ##v_0=12 m/s## and ##v=0 m/s##.
 
  • #15
PeroK
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I am supposed to use integration. I am unable to use an equation of motion using any three variables. I have ##v_0=12 m/s## and ##v=0 m/s##.
So, what precisely is stopping you?

It's not too hard. My suggestion is to simplify things by defining:$$a(T) = -cT \ \ \ (0 < T < T_2)$$ where ##T = t - t_1## That lets you calculate ##T_2## and ##x(T_2)## more easily. Then just add on the initial distance before the deceleration phase.
 
  • #18
PeroK
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The first thing you need to do is get an expression for ##v(T)## (or ##v(t)##, if you don't like my approach).

Can you do that?

It's difficult to help if we can't see where you're having the problem.
 
  • #19
I am just having a problem in 2
The first thing you need to do is get an expression for ##v(T)## (or ##v(t)##, if you don't like my approach).

Can you do that?

It's difficult to help if we can't see where you're having the problem.
I am just having a problem in two things, namely:
1. How to find the difference ##t_2-t_1## to find the acceleration using equations of motion, or finding the acceleration directly using the given constant ##c##, and
2. Where and how to use the constant ##c##(for integration) .
 
  • #20
I think it will be easy to use the equations of motion rather than use calculus.
 
  • #21
PeroK
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I think it will be easy to use the equations of motion rather than use calculus.
If you mean SUVAT, they are only valid for constant acceleration. There's no alternative to calculus in this case.
 
  • #22
jbriggs444
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find the acceleration using equations of motion
The acceleration is given in the problem statement. You don't need to compute it.
 
  • #23
If you mean SUVAT, they are only valid for constant acceleration. There's no alternative to calculus in this case.
Oh yeah, you are right.
 
  • #24
The acceleration is given in the problem statement. You don't need to compute it.
Yes, you are right. I integrated ##f(a)## for the second phase by putting the values and got a quadratic equation and then I solved it and got ##t_2 = 3s##.
 
  • #25
Also, I integrated the speed function (integrated ##f(a)## twice) and got ##x_{t_2} = 10 m##. How is that possible since the car is stopping while going forward and so ##x_{t_2}>12m##.
 
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