Finding equation of the position

[Lord Dark]

Homework Statement

hi everyone ,, this is my last question before exam :

1) The figure shows the velocity versus time graph. If a particle starts motion at t=0.0
from x0 = 10.0 m:
a) Write the equation describing the position of the particle as a function of time t.
b) What are the positions of the particle at t= 2 s and t = 4 s.?
c) What is the position of the particle at the instant when it momentarily stops?
d) Find the distance and displacement of the particle between t = 2 s and t = 5 s.

Homework Equations

The Attempt at a Solution

I know that a=0 ,, and at t=0 > x=10m and at the same time Vo=12 m/s ,, and the velocity is decreasing so I made this equation :x=10-12t but the problem at t=0 Vo=0 and when I apply x(3) I don't get 18m which it's what i have to get by the graph .. can someone tell me what to do to get the equation ..?

 

[LowlyPion]

Consider how you find displacement from a velocity time graph.

If V = dX/dt, then isn't

X = ∫ V dt ?

And if you integrate V on your graph, what is the easiest way to do that?

Can you think of something that rhymes with area?

 

[Lord Dark]

Consider how you find displacement from a velocity time graph.

If V = dX/dt, then isn't

X = ∫ V dt ?

And if you integrate V on your graph, what is the easiest way to do that?

Can you think of something that rhymes with area?

I didn't understand you quite well ,, I tried by doing X = ∫ V dt from x(0,3) x=18 and i get v=6 m/s ,, and from x(0,5) v=2 m/s but the problem is at x(0,0) I should get x=10 and I don't get it,, V changes ...

 

[LowlyPion]

I didn't understand you quite well ,, I tried by doing X = ∫ V dt from x(0,3) x=18 and i get v=6 m/s ,, and from x(0,5) v=2 m/s but the problem is at x(0,0) I should get x=10 and I don't get it,, V changes ...

Xo is the constant of integration isn't it?

 

[Lord Dark]

yea Xo is constant ,, should I include it in the integral or X = Xo + ∫ V dt ??
and BTW when i integrate ∫ V dt ,, I'll get Vt right ??

 

[LowlyPion]

yea Xo is constant ,, should I include it in the integral or X = Xo + ∫ V dt ??

V_(t) = 12 - 4t isn't it?

So ∫ V_(t) dt = ∫ (12 - 4t) dt = X_(t)

That means that X_(t) = 12t - 1/2*4*t² + C

Where C = X_o

 

[Lord Dark]

aha ,, so that what you meant X = ∫ V ,, I didn't know that first I should get V(t) equation then integrate .. thanks very much :D

 

[LowlyPion]

aha ,, so that what you meant X = ∫ V ,, I didn't know that first I should get V(t) equation then integrate .. thanks very much :D

You can also solve through visual inspection, by simply calculating the area under the curve. Each square represents 4 m.