Velocity/Accelration/Displacment all in one question.

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SUMMARY

The discussion revolves around solving kinematic problems involving velocity, acceleration, and displacement using a velocity versus time graph. Participants analyze how to derive equations for velocity segments, calculate acceleration, and determine the area under the curve to find displacement. Key equations discussed include the integration of velocity to find position and the use of slopes to determine acceleration. The final position and speed at specific times are calculated based on the graph's segments.

PREREQUISITES
  • Understanding of kinematics and motion equations
  • Familiarity with integration techniques in calculus
  • Ability to interpret and sketch velocity versus time graphs
  • Knowledge of calculating areas under curves for displacement
NEXT STEPS
  • Learn how to derive equations for velocity segments from graphs
  • Study integration methods for calculating displacement from velocity
  • Explore the relationship between acceleration and the slope of velocity graphs
  • Practice solving kinematic problems using real-world scenarios
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of graph interpretation and problem-solving techniques in motion analysis.

  • #31
A particle starts from rest and accelerates as shown in Figure P2.11. The divisions along the horizontal axis represent 3.0 s and the divisions along the vertical axis represent 2.0 m/s2.
 
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  • #32
This part should be easier, because it doesn't ask you to plot the graph... yeah, it's similar in concept to the previous one...

What is the area under an a-t graph?
 
  • #33
it should be the v(t).
 
  • #34
veloix said:
it should be the v(t).

yes... generally it would be the change in v(t)... but since here it starts from rest it is v(t) itself...

So what is v(t) at t = 6.0s?
 
  • #35
v(t) at t= 6.0s should be ∫v(t)dt
 
  • #36
veloix said:
v(t) at t= 6.0s should be ∫v(t)dt

Yes, you can just get the area under the curve.
 
  • #37
when i intagrate that i get t^2/2?
 
  • #38
veloix said:
when i intagrate that i get t^2/2?

I don't understand. did you get the velocity at t = 6?

What's the area under the a-t graph from 0 to 6? Just get the area of the rectangle.
 
  • #39
yea it comes out to be 18m/s but it wrong answer.
 
  • #40
veloix said:
yea it comes out to be 18m/s but it wrong answer.

how did you get 18?
 
  • #41
t= 6 v(t)=6^2/2=18m/s
 
  • #42
the area of the rectanlge would 6X2=12
 
  • #43
veloix said:
t= 6 v(t)=6^2/2=18m/s

The a-t graph is just a rectangle... area of a regtangle is length times width = 6*4 = 24.
 
  • #44
oh that's why i got it wrong i had 2 written on my y-axis instead of 4, ugg. so to get the speed at 12s i would get area of each rectangle and add them up.
 
  • #45
veloix said:
oh that's why i got it wrong i had 2 written on my y-axis instead of 4, ugg. so to get the speed at 12s i would get area of each rectangle and add them up.

careful about the 12s though... first get v(12)... take the area above the x-axis, then subtract the area below the x-axis...

That gives v(12), which may be negative... so the speed would be the absolute value of that.
 
  • #46
great that work out perfectly thank you
part c asking for distance this time, the ∫d(t)?
 
  • #47
veloix said:
great that work out perfectly thank you
part c asking for distance this time, the ∫d(t)?

Yeah, plot v(t)... then get the area... you'll need to add all the areas (ie don't subtract areas under the x-axis, just add them)... because the question asks for distance, not displacement...
 
  • #48
i can't draw this plot right I am haveing difficluty.
 
  • #49
veloix said:
i can't draw this plot right I am haveing difficluty.

you should have 3 lines... positive slope... 0 slope (ie horizontal)... and negative slope at the end...
 
  • #50
oh yea that makes sense ty.
 

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