Finding the magnitude of a particle from a graph?

In summary, the conversation discusses calculating displacement, distance, and average velocity for a particle on a straight line with given initial position and velocity. The equations x - x0 = v0t + (1/2)at2 and v = v0 + at are used to find the displacement and velocity at different time intervals. The average velocity is the total displacement divided by the time period. Care must be taken with the endpoints when the acceleration is discontinuous.
  • #1
themadcow45
11
0

Homework Statement



Consider the acceleration of a particle along a straight line with an initial position of 0 m and an initial velocity of 0 m/s.

5b9pgo.png


1) Calculate the magnitude of the displacement after the car travels the first 5 s.
Answer in units of m.

2) How far does the car travel from 9 s to 13 s, if it does not experience any acceleration during this time period?
Answer in units of m.

3) Calculate the magnitude of the car’s average velocity from 5 s to 9 s.
Answer in units of m/s.

Homework Equations



x - x0 = v0t + (1/2)at2

The Attempt at a Solution



For 1), I guess I'm confused at how to be reading the graph. I've tried breaking it up into 0 to 1 s, 1 to 3 s, 3 to 4 s, and 4 to 5 s, and using the equation above to find the displacement for each interval, but I'm getting confused as to where the acceleration is located on the graph. For example, is the acceleration at t = 3 s supposed to be 2 m/s2 or 0 m/s2?

For 2), when it says that it doesn't experience any acceleration, does that mean acceleration = 0?

For 3), what is the equation for average velocity?

Any help/hints would be appreciated, thanks!
 
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  • #2
themadcow45 said:
For 1), I guess I'm confused at how to be reading the graph. I've tried breaking it up into 0 to 1 s, 1 to 3 s, 3 to 4 s, and 4 to 5 s, and using the equation above to find the displacement for each interval, but I'm getting confused as to where the acceleration is located on the graph. For example, is the acceleration at t = 3 s supposed to be 2 m/s2 or 0 m/s2?
Consider thinking in terms of open intervals (mathematically speaking) when the endpoints are discontinuous. So intervals might be:

0 ≤ t < 1
1 ≤ t < 3
3 ≤ t < 4
4 ≤ t < 5
5 ≤ t < 6

and so on.

The end-point value that you use in your equations for any interval should reflect the value that the variable had within that interval immediately adjacent to the endpoint.
For 2), when it says that it doesn't experience any acceleration, does that mean acceleration = 0?
Yes.
For 3), what is the equation for average velocity?
Average velocity is defined to be the total displacement divided by the time.
 
  • #3
Hi gneill,

So for 1), here's my work:

0 to 1: x - 0 = 0(1) + .5(0)(1)^2 = 0
1 to 3: x - 0 = 0(2) + .5(2)(2)^2 = 4
3 to 4: x - 4 = 4(1) + .5(1)(1)^2 = 8.5
4 to 5: x - 8.5 = 4(1) + .5(1)(1)^2 = 13

Would 13 be the correct answer, or am I missing anything?
 
  • #4
Between 3 and 4 seconds the acceleration should be zero. So only the current velocity (from the end of the last stage) should be affecting the distance.

You can move the initial position values to the RHS of your equations so that the equations make algebraic sense for the final equals sign. So for any given step:

xnew = xp + vp*Δt + (1/2)*a*(Δt)2
vnew = vp + a*Δt

where the "p" subscript refers to the value at the end of the previous step.
 
  • #5
I've gotten everything except the last problem;

3) Calculate the magnitude of the car’s average velocity from 5 s to 9 s.
Answer in units of m/s.

Here's my work so far:
5 to 6: v = 13, x = 21.5
6 to 8: v = 11, x = 45.5
8 to 9: v = 3, x = 52.5

You said the average velocity is total displacement divided by the time, so it would be:

(21.5 + 45.5 + 52.5) / 4, right? But that answer is coming out as wrong...does anyone know what I'm doing wrong?

I also tried (13 + 3) / 2, but that's wrong as well.
 
  • #6
You want the distance traveled during the specified time period, divided by the length of the time period.

What is the displacement at the beginning of second 5? At the end of second 9?
 
  • #7
Oh...I think I understand...So it would be (21.5 + 52.5) / 4 ?
 
  • #8
themadcow45 said:
Oh...I think I understand...So it would be (21.5 + 52.5) / 4 ?

Is 21.5 the displacement at the beginning of second 5 (same as end of second 4)?
 
  • #9
Ah, I read your previous post wrong, my fault... it would be 12.5.

But would 52.5 still be the end point, right, since you can't extend the graph to t = 10?
 
  • #10
themadcow45 said:
Ah, I read your previous post wrong, my fault... it would be 12.5.

But would 52.5 still be the end point, right, since you can't extend the graph to t = 10?

Correct.
 
  • #11
gneill said:
Correct.

(12.5 + 52.5) / 4 is coming out to be incorrect...Any tips on what I'm doing wrong?
 
  • #12
themadcow45 said:
(12.5 + 52.5) / 4 is coming out to be incorrect...Any tips on what I'm doing wrong?

Total displacement is the final position minus the initial position. Don't sum them!
 
  • #13
gneill said:
Total displacement is the final position minus the initial position. Don't sum them!

D'oh. Physics isn't my thing, but thanks for the help gneill - you've been invaluable in your patience and quick responses. Thank you!
 

1. What is the definition of magnitude in physics?

Magnitude in physics refers to the size or quantity of a physical property. It can be measured in various units depending on the specific property being described.

2. How is magnitude represented on a graph?

Magnitude is typically represented by the length of a vector on a graph. The longer the vector, the greater the magnitude.

3. What is the process for finding the magnitude of a particle from a graph?

To find the magnitude of a particle from a graph, you would need to measure the length of the vector representing the particle's position. This can be done using a ruler or by using the scale provided on the graph. The length of the vector is the magnitude of the particle's position.

4. Can the magnitude of a particle change over time?

Yes, the magnitude of a particle can change over time as its position changes. For example, if a particle is moving in a straight line, its magnitude will increase as it moves further away from its starting point.

5. How can the magnitude of a particle be used in physics calculations?

The magnitude of a particle can be used in various physics calculations, such as determining the speed or velocity of the particle, calculating its acceleration, or finding the net force acting on the particle. It is an important factor in many equations and can provide valuable information about the motion of a particle.

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