Velocity/Accelration/Displacment all in one question.

[veloix]

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.

 

[learningphysics]

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?

 

[veloix]

it should be the v(t).

 

[learningphysics]

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?

 

[veloix]

v(t) at t= 6.0s should be ∫v(t)dt

 

[learningphysics]

v(t) at t= 6.0s should be ∫v(t)dt

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

 

[veloix]

when i intagrate that i get t^2/2?

 

[learningphysics]

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.

 

[veloix]

yea it comes out to be 18m/s but it wrong answer.

 

[learningphysics]

yea it comes out to be 18m/s but it wrong answer.

how did you get 18?

 

[veloix]

t= 6 v(t)=6^2/2=18m/s

 

[veloix]

the area of the rectanlge would 6X2=12

 

[learningphysics]

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.

 

[veloix]

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.

 

[learningphysics]

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.

 

[veloix]

great that work out perfectly thank you
part c asking for distance this time, the ∫d(t)?

 

[learningphysics]

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...

 

[veloix]

i can't draw this plot right I am haveing difficluty.

 

[learningphysics]

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...

 

[veloix]

oh yea that makes sense ty.