# Homework Help: A plot of acceleration versus time for a particle is shown

1. Sep 13, 2007

### suspenc3

1. The problem statement, all variables and given/known data
A plot of acceleration versus time for a particle is shown. Whats the difference between its position at t=4s and t=0s if v(0)=-4m/s.

http://img441.imageshack.us/img441/532/dynmc8.th.png [Broken]

2. Relevant equations

3. The attempt at a solution
Ive been working on this problem for hours until I thought I solved it...I got the right answer but I realised after that I made a mistake..but still got the right answer. After I tried to do it the right way i keep getting the wrong answer.

v(0)=-4m/s
v(1)=16m/s (since constant acceleration from 1s to 2s)

$$a_{12} = -20t+40 = dv/dt$$ - y=mx+b, for the non-uniform section (1s-2s)

now I rearrange the above and integrated to find v(2)

$$\int_{v(1)}^{v(2)}dv = \int_1^2 -20t+40dt$$

This is where I made my mistake, instead of using v(1) I used v(0)=-4m/s

continuing...$$v(2)-(-4m/s)=-10t^2+40t |_1^2$$

solving gives v(2)=26m/s

Now I integrated again...$$-10t^2+40t = dx/dt$$ to find the displacement for the non-uniform section...giving me an answer of 68/3.

Anyways..add all the displacements up and I end up with 80.66 wich is the right answer. How come when I try and use v(1)=16m/s for my lower limit in the first integration I keep getting wrong numbers?

Any help would be appreciated

Last edited by a moderator: May 3, 2017
2. Sep 16, 2007

### Astronuc

Staff Emeritus
It is difficult to read the variable on the vertical axis of the plot, which looks like x, or position, rather than acceleration.

If the plot is acceleration, and the vertical axis is at t=0, then the acceleration is constant 20 m/s2 through t = 1s, then it decreases linearly between 1 and 2 to zero, after which there is no acceleration from t = 2s to 4s (so for the last two seconds, it's coasting at constant velocity).

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