Velocity/Accelration/Displacment all in one question.

  • Thread starter veloix
  • Start date
In summary: The slope of the line is the acceleration... the area between the line and the x-axis is the distance travelled... the slope of the line gives you the acceleration and the area between the line and the x-axis gives you the distance travelled.x(t)= (1/2)(at^2) + v_0t + x_0a is the slope of the line... ie the acceleration.v_0 is the initial velocity at t = 0...x_0 is the initial position at t = 0.In summary, the conversation discusses a homework problem involving a velocity versus time graph of a moped traveling along a straight road. The graph shows a positive slope of 14 m/s for the first
  • #36
veloix said:
v(t) at t= 6.0s should be ∫v(t)dt

Yes, you can just get the area under the curve.
 
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  • #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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