# Homework Help: Velocity/Accelration/Displacment all in one question.

1. Sep 19, 2007

### veloix

1. The problem statement, all variables and given/known data

A student drives a moped along a straight road as described by the velocity versus time graph in Figure P2.14. The divisions along the horizontal axis represent 1.5 s and the divisions along the vertical axis represent 3.5 m/s.
Sketch this graph in the middle of a sheet of graph paper. (Do this on paper. Your instructor may ask you to turn in your work.)

Figure P2.14
<img src=http://a425.ac-images.myspacecdn.com/images01/103/l_3fca2200153cb8eb21ea540316125000.gif>
(a) Directly above your graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (Do this on paper. Your instructor may ask you to turn in your work.)

(b) Sketch a graph of the acceleration versus time directly below the vx-t graph, again aligning the time coordinates. (Do this on paper. Your instructor may ask you to turn in your work.)
On each graph, show the numerical values of x and ax for all points of inflection.

(c) What is the acceleration at t = 9.0 s?
m/s2
(d) Find the position (relative to the starting point) at t = 9.0 s.

m
(e) What is the moped's final position at t = 13.5 s?
m

2. Relevant equations
all of the kinematics and equations dealing with velocity,time,positon,accerlation
will help out

3. The attempt at a solution

let see i figured the part looking for accelration out by taking the a tanget of the line near 9s using equation for slope form. vf-vi/tf-ti= 0-14m/s/12s-9s=-4.67m/s^2

now the rest im totaly lost ive tried to intergrate the function of the graph but im not sure how to come up with the equation of the function. can anyone help me out?

2. Sep 19, 2007

### veloix

heres the graph

3. Sep 19, 2007

### learningphysics

The acceleration is just the slope of the lines...

The change in position is the area under the v-t graph...

4. Sep 19, 2007

### veloix

ok the area under the curve, now this where im getting stuck. do i use intergration?

5. Sep 19, 2007

### learningphysics

You don't have to... suppose I'm looking at the first segment of the curve (the positive slope part)... so 0 up to some time t... what is the area under the graph ?

6. Sep 19, 2007

### veloix

lets see the area under the first part. looks like two triangles. so area of the first triangle is bXhX1/2 =1/2 4.5X(14)=31.5 and the 2nd 1/2X4.5X(14)=31.5 so 31.5+31.5= 63

Last edited: Sep 19, 2007
7. Sep 19, 2007

### learningphysics

Why are you getting 2 triangles? I just see one...

suppose you aren't going all the way to t = 4.5. you're calculating the area from t = 0... to some t = tf... what is the area of the triangle?

Actually maybe it is better to take the integral... so for the first segment... what is the equation for v(t) ? ie: for t = 0 to t = 4.5...

8. Sep 19, 2007

### veloix

equation for v(t)= absx

9. Sep 19, 2007

### learningphysics

what is abs?

You want an equation for v(t) in terms of time...

10. Sep 19, 2007

### veloix

v(t)=│t│
t=0
v(t)=│0│=0
t=4.5
v(t)=│4.5│=4.5

Last edited: Sep 19, 2007
11. Sep 19, 2007

### learningphysics

Hmmm... how do you get that? I see the slope as 14/4.5 = 3.111

12. Sep 19, 2007

### learningphysics

Isn't v = 14 at t = 4.5?

You don't need to use absolute value... since time is positive anyway.

13. Sep 19, 2007

### veloix

yea v= 14 at t= 4.5 , my mistake on that.

Last edited: Sep 19, 2007
14. Sep 19, 2007

### veloix

so if i were to intergrate that. it be ∫v(t) dt

Last edited: Sep 19, 2007
15. Sep 19, 2007

### learningphysics

dt not dx.

But that equation for v(t) only holds for t = 0 to t = 4.5... What is the equation of each segment of v(t) ?

When you get that, then you can get the integrals...

16. Sep 19, 2007

### veloix

t=0 to t= 4.5
v(t)=t
t=4.5 to t=9
v(t)=14
t=9 to t= 15
v(t)=(-t-12)

17. Sep 19, 2007

### learningphysics

But the slope of the line is 14/4.5 = 3.11 so this can't be right.

Yes, good.

you calculated the slope of this part in your original post... it's not -1

18. Sep 19, 2007

### veloix

ok so first one must be v(t)=3.11t
and the last is v(t)=(-4.67t-12)

19. Sep 19, 2007

### learningphysics

yes.

The slope is right, but the y-intercept is wrong.

20. Sep 19, 2007

### veloix

hmm well i figured it was a reflection of the f(t) and it shifted to the right to 12. but i guess i didnt do it right. should i extend the line to the y axis to get the intercept?

21. Sep 19, 2007

### veloix

ok shoot now i c. it should be v(t)=(-4.67t-52.5)

22. Sep 19, 2007

### learningphysics

The y-intersepct should be positive... the line segment when extended will hit the y-axis at a positive value.

you know that v(t) =-4.67t + b, just plug in the t value and v value at a point... then solve for b.

23. Sep 19, 2007

### veloix

ok thanks

24. Sep 19, 2007

### veloix

(v)t=(-4.67t+56.04)

25. Sep 19, 2007

### learningphysics

I'm getting v(t) = -4.67t + 49.03, using v(10.5) = 0