# Determine the signs (positive or negative) of the particle

1. Sep 13, 2015

### wein7145

Background
The teacher doesn't teach we just do activities and have whole chapters worth of 20 mini assignments due every few days. Yes I did read the book.

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

Determine the signs (positive or negative) of the position, velocity, and acceleration for the particle in
FIGURE Q1.6, FiGURE Q1.7 and FIGURE 1.8:

Figure 1.6

Figure 1.7 and Figure 1.8

2. Relevant equations
x=x(t) distance or position

__dx
v= -- velocity
__dt

__dv_____d^(2)x
a= -- or --------- acceleration
__dt_____dt^(2)

Ignore this "__" it is a space this doesn't accept spaces like some languages do...

3. The attempt at a solution
For 1.6 I'm pretty sure Position is (negative) from starting point to origin
For 1.7 I'm pretty sure Position is (positive) from starting point to origin
For 1.8 I'm pretty sure Position is (negative) from starting point to origin

How can you determine velocity and acceleration which are derivatives from a line that has no equation or labeled points??? I'm also aware of what is called inflection points where it changes from positive to negative. Are there rules I'm missing here? Would all signs be the same since this is 1D motion??

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2. Sep 13, 2015

### JorisL

You can determine the velocity from the direction of the arrows.
The acceleration is related to the length of the arrows.

You mixed up the answers for 1.7 and 1.8
Remember that the y-axis is generally pointing upwards.

3. Sep 13, 2015

### wein7145

Thanks that helps a lot
I thought it was just showing the velocity only not acceleration. Because the previous ones showed velocity with a direction bar ("v-bar") but looking at it again there are no markings. Yes so in conclusion: (That also makes sense from a number line point of view as -5 to -1 is still negative and 1 to 5 is still positive)
For 1.6 I'm sure Position is (negative) from starting point to origin
For 1.7 I'm sure Position is (negative) from starting point to origin (in math terms this is more positive but still negative)
For 1.8 I'm sure Position is (positive) from starting point to origin

4. Sep 13, 2015

### JorisL

The position is strictly negative whatever way you look at it.
You being confused by the arrows.
The arrows point in the positive direction for 1.7, this means the velocity is positive.

Can you figure out the other two velocities?

Remember, $a=\frac{dv}{dt}$ so you can look at how the velocity changes to determine the acceleration.

5. Sep 13, 2015

### wein7145

For 1.6 I'm sure Position is (negative), velocity is (positive), acceleration is (negative) from starting point to origin
For 1.7 I'm sure Position is (negative), velocity is (positive), acceleration is (positive) from starting point to origin
For 1.8 I'm sure Position is (negative), velocity is (positive), acceleration is (positive) from starting point to origin
Assuming all signs the same rule. Unless this is asking for signs at each part increasing and decreasing. Now I'm thinking the small arrow is decreasing and the larger arrow is increasing. Unless all arrows are just increasing? Again I don't even know the rules but this may in the end finally teach me them...

Whoops. Hopefully you can see the edits.

Last edited: Sep 13, 2015
6. Sep 13, 2015

### JorisL

Almost there, you had the positions right in your previous post (negative, negative and positive). I was aiming at the small remark you made.
Why do you think the acceleration is negative in the first picture?
Doesn't the velocity grow?

7. Sep 13, 2015

### wein7145

Because in terms of derivatives usually they do the opposite sign at the inflection point.
For 1.6 I'm now sure Position is (negative), velocity is (positive), acceleration is (positive) from starting point to origin
For 1.7 I'm now sure Position is (negative), velocity is (positive), acceleration is (positive) from starting point to origin
For 1.8 I'm now sure Position is (postive), velocity is (positive), acceleration is (positive) from starting point to origin

8. Sep 13, 2015

### JorisL

About the signs of acceleration, if the velocity increases you know that the acceleration has the same sign as this velocity.

You can easily see this from the kinematic equation $v(t) = v_0 + a\cdot t$ where I defined $v_0$ as the initial velocity.
.

As an aside, it's better to use speed when discussing the size of the velocity.
This helps you remember this distinction and avoid mistakes along the line.

Good Luck

J

9. Sep 13, 2015

### wein7145

Yes I agree speed and graphs with slope describe this concept better. (Although technically the slope is zero)

Going through my mistakes helps me remember and avoid making the same mistakes again. Thank you for the help. Strange how the book doesn't have answers to these conceptual questions but has answers to most of the odd numbered questions. I will have many more questions so be on the lookout for my journey into Physics! :)