Velocity/Accelration/Displacment all in one question.

[veloix]

Homework Statement

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

Homework Equations

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

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 I am totally lost I've tried to intergrate the function of the graph but I am not sure how to come up with the equation of the function. can anyone help me out?

 

[veloix]

heres the graph

 

[learningphysics]

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

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

 

[veloix]

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

 

[learningphysics]

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

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 ?

 

[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

 

[learningphysics]

lets see the area under the first part. looks like two triangles. so area of the first triangle is bX1/2h = 4.5X1/2(14)=31.5 and the 2nd 4.5X1/2(14)=31.5 so 31.5+31.5= 63

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...

 

[veloix]

equation for v(t)= absx

 

[learningphysics]

equation for v(t)= absx

what is abs?

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

 

[veloix]

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

 

[learningphysics]

sry ment absoultevalue of t.

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

 

[learningphysics]

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

Isn't v = 14 at t = 4.5?

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

 

[veloix]

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

 

[veloix]

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

 

[learningphysics]

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

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...

 

[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)

 

[learningphysics]

t=0 to t= 4.5
v(t)=t

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

t=4.5 to t=9
v(t)=14

Yes, good.

t=9 to t= 15
v(t)=(-t-12)

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

 

[veloix]

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

 

[learningphysics]

ok so first one must be v(t)=3.11t

yes.

and the last is v(t)=(-4.67t-12)

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

 

[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?

 

[veloix]

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

 

[learningphysics]

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

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.

 

[veloix]

ok thanks

 

[veloix]

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

 

[learningphysics]

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

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

 

[veloix]

i c yes you are right i was wrong again. v(t)=-4.67+49.03 using v(10.5)=0

 

[learningphysics]

Anyway, from here you can calculate the position using:

$$x(t)= x_0 + \int_0^{t}v(t)dt$$

I think perhaps x_0 = 0 ?

so:

$$x(t)= \int_0^{t}v(t)dt$$

you'll have 3 sections: where t = 0 to t = 4.5... from t=4.5 to t=7.5, and from t=7.5 to 13.5

ah... you had 9, instead of 7.5... maybe that was where the mistake was coming from.

 

[veloix]

ok I am going try to work this out

 

[learningphysics]

ok I am going try to work this out

cool. when you take the integral from 0 to some value between 4.5 and 7.5... eg 6... you split the integral into two parts... the integral from 0 to 4.5, and the integral from 4.5 to 6... ie divide them into the segments... hope that makes sense.

 

[veloix]

wow i got the right answer. you were a very big help on this one i don't know how i can thank you but there is one more problem i have. let me show this graph for a sec.

(a) Determine the particle's speed at t = 6.0 s.


What is the speed at t = 12.0 s?


(b) Find the distance traveled in the first 12.0 s.
m
is this going be another one like the prevoius one? when the lines are going down like that one the graph, i not sure what the velocity is doing i know when velocity is changing accelration will be changing. and when the velocity is is not changeing accelration is 0.