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A glider of length 12.4 cm moves on an air track

  • #1

Homework Statement


Serway Physics - P2.40

A glider of length 12.4 cm moves on an air track with constant acceleration (Fig P2.39).
A time interval of 0.628 s elapses between the moment when its front end passes a fixed
point A along the track and the moment when its back end passes this pont. Next, a time
interval of 1.39 s elapses between the moment when the back end of the glider passes the
point A and the moment when the front end of the glider passes a second point B further
down the track. After that an additional 0.431 s elapses until the back end of the glider
passes point B.
(a) Find the average speed of the glider as it passes point A.
(b) Find the acceleration of the glider.
(c) Explain how you can compute the acceleration without knowing the distance
between points A and B

Homework Equations




The Attempt at a Solution


[/B]
(a) For constant acceleration I know that ##v_{x,avg}=\frac {v_{xi} + v_{xf}} {2} ## (1)

Also I know that ##x_f=x_i + \frac {v_{xi} + v_{xf}} {2} t## (2)

So ##(2) => v_{x,avg} = \frac {x_f - x_i} {t} = \frac {12.4cm} {0.626s} = 19.74cm/s##
Is part (a) correct?

How do I find part (b) and (c)?
 

Answers and Replies

  • #2
kuruman
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Part (a) is correct for the average speed. To do parts (b) and (c) you need to bring in another "relevant" equation. You are given times and positions of the glider. What could that equation be?
 
  • #3
Is it another equation under constant acceleration section of the book?
or you mean I should bring a velocity equation (since it is related with position and time) ?
 
  • #5
Do you mean this equation: ##x=x_0+v_0 t+\frac {1} {2} at^2## ?
 
  • #6
kuruman
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That's the one.
 
  • #7
Well, I am wondering how to use this equation...
What is ##v_0##? Do I know ##v_0##?
 
  • #8
kuruman
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Well, I am wondering how to use this equation...
What is ##v_0##? Do I know ##v_0##?
You don't but you have information to write two equations and two unknowns, ##a## and ##v_0##. Use them to eliminate ##v_0## and find ##a##.
 
  • #9
Is the following correct? (for part b)

Serway P2_40001.jpg
 
  • #10
kuruman
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The first equation is correct but the second equation is not. You use the symbol ##v_0## to denote the velocity when the front end of the glider crosses point A. Because the glider accelerates, the velocity will no longer have that value when the glider crosses point B. You need to connect with an equation the velocity of the glider when its leading edge crosses point A, call it ##v_{0A}##, to the velocity of the glider when its leading edge crosses point B, call it ##v_{0B}.##
 
Last edited:
  • #11
The first equation is correct but the second equation is not. You use the symbol ##v_0## to denote the velocity when the front end of the glider crosses point A. Because the glider accelerates, the velocity will no longer have that value when the glider crosses point B. You need to connect with an equation the velocity of the glider when its leading edge crosses point A, call it ##v_{0A}##, to the velocity of the glider when its leading edge crosses point B, call it ##v_{0B}.##
Hi! Do you mean this equation ##v_{0B} = v_{0A} + at## where t = 0.628s + 1.39s ?
 
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  • #12
kuruman
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That's exactly what I mean!
 
  • #13
jbriggs444
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Personally, I would attack this by noting that, for constant acceleration, the velocity at the mid-point of a time interval is equal to the average velocity over that time interval.

That fact seems intuitively obvious, but can be derived if needed:

$$v_{ave} = \frac{v_i + v_f}{2}$$
$$= \frac{v_i + (v_i + a \Delta t)}{2}$$
$$= \frac{(v_i + a \Delta t/2) + (v_i + a\Delta t/2)}{2}$$
$$=\frac{2(v_i + a \Delta t/2)}{2}$$
$$=v_i + a \Delta t/2$$
$$=v_{mid time}$$

This clears the way for a very straightforward calculation.
 
  • #14
kuruman
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Personally, I would attack this by noting that, for constant acceleration, the velocity at the mid-point of a time interval is equal to the average velocity over that time interval.
Yes indeed, I was saving that approach for later, after OP completed the problem. The second approach makes the algebra simpler while the first approach has an easier-to-understand strategy and setup.
 
  • #15
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@J.Briggs - you beat me to it! Was just checking result on Wolfram Alpha as compared to OP's calculations as corrected by Kuruman. This is a very useful method to employ for (uniformly accelerating) projectiles passing a given space 'window' in a given time 'window'.
 

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