Distance and displacement confusion

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Homework Help Overview

The discussion revolves around two objects moving with initial velocities of -8.00 m/s and final velocities of 16.0 m/s, both experiencing constant acceleration. The first object has a displacement of 20.0 m, while the second travels a total distance of 22.0 m. Participants are attempting to find the accelerations of both objects using the kinematic equation Vf^2 = Vi^2 + 2ax.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are discussing the application of the kinematic equation and questioning the initial and final velocities used in the calculations. Some express confusion about the displacement values and the direction of acceleration. There are attempts to split the motion into two phases to analyze the acceleration, with questions about how to relate time and acceleration.

Discussion Status

The discussion is ongoing, with multiple interpretations being explored regarding the relationship between time, distance, and acceleration. Some participants have provided guidance on considering the motion in two parts, while others are questioning the assumptions made about the distances and accelerations.

Contextual Notes

Participants are grappling with the implications of constant acceleration and how it affects the time and distance traveled during different phases of motion. There is a lack of consensus on the correct approach to relate the various quantities involved in the problem.

Ab17
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1. The problem statement, all variables and given/known

Two objects move with initial velocity -8.00 m/s xfinal velocity 16.0 m/s xand constant accelerations. (a) The first object has displacement 20.0 m. Find its accelera- tion. (b) The second object travels a total distance of 22.0 m. Find its acceleration. x

2. Homework Equations

Vf^2=vi^2 + 2ax


The Attempt at a Solution


(a) (-8)^2 = (16)^2 + 2a(22)
a = 4.36m/s

(b) 0 = (-8)^2+ 2a(11)
a= -2.91m/s2

(16)^2= (0)^2 + 2a(11)
a= 0.73m/s2

a = -2.91 +0.73
a= -2.18m/s2
 
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Is this how it is done?
 
Ab17 said:
1. The problem statement, all variables and given/known

Two objects move with initial velocity -8.00 m/s xfinal velocity 16.0 m/s xand constant accelerations. (a) The first object has displacement 20.0 m. Find its accelera- tion. (b) The second object travels a total distance of 22.0 m. Find its acceleration. x

2. Homework Equations

Vf^2=vi^2 + 2ax


The Attempt at a Solution


(a) (-8)^2 = (16)^2 + 2a(22)
a = 4.36m/s
You've got your Vf and Vi mixed up...and displacement is 20 m , not 22 m...what's the direction of the acceleration?
(b) 0 = (-8)^2+ 2a(11)
a= -2.91m/s2

(16)^2= (0)^2 + 2a(11)
a= 0.73m/s2

a = -2.91 +0.73
a= -2.18m/s2
the acceleration is constant, so you can't have 2 values. You incorrectly assumed it goes 11 m each way. The distance it travels from rest varies as the square of the final speed.
 
Can you please explain fully how part b is done I don't understand
 
Ab17 said:
Can you please explain fully how part b is done I don't understand
For part b, consider splitting the trajectory into two parts. One as the object slows from -8 m/s to zero. And one as the object accelerates from zero to +16m/s. How long does the second part last compared to the first?
 
Thats what I done. I got two accelerations
 
Ab17 said:
Thats what I done. I got two accelerations
See post #3. You split it in half. Why half? You are told that the acceleration is constant. That information allows you to determine what fraction of the total time elapses in each part.
 
The question asks for acceleration and not time
 
Can you please explain by an example. I don't understand what you saying
 
  • #10
Ab17 said:
Can you please explain by an example. I don't understand what you saying
The problem says that the acceleration, whatever it is, is constant. So take it as a constant and call it "a".

How much time elapses while the object slows down from -8 m/s to 0 m/s?
How much time elapses while the object speeds up from 0 m/s to 16 m/s?
Express both answers in terms of a.

Edit: let me expound a bit on the strategy that I have in mind...

In general terms, we are looking at an object that is moving first to the left and then to the right under a constant rightward acceleration.

We want to know the acceleration, a. We know the starting velocity, ending velocity, total distance covered and the fact that the acceleration is constant. Ideally we could find a single equation involving starting velocity, ending velocity, distance moved along the path and acceleration and solve for acceleration.

We have an equation that relates starting velocity, ending velocity, displacement from start to end and acceleration. But that equation does not help much because the path folds back along itself. The distance moved along the path does not match the displacement from start to end.

If we can split the path into two parts, each of which involves a straight line path that never reverses itself then we can use that displacement equation for each part separately. There are a number of ways to use this general approach. The one I have in mind is...

Working backward, the hope is that we can figure out what fraction of the total distance is covered in the first (or second -- it does not matter which) part of the event. If we knew either one, we'd be in a position to solve for the required acceleration.

How do we determine the fraction of the total distance covered? If we knew the fraction of the total time elapsed for each part, that information might come in handy.

So, what fraction of the total time is required for each part? That motivates the questions I've asked above.
 
Last edited:
  • #11
Vf = vi + at
0 = -8 + at
a = 8/t

16= 0 +at
a = 16/t
 
  • #12
That does not answer the questions that I posed.
 
  • #13
Xf = xi + 1/2(Vf + Vi)t
-11 = 0.5(0 -8)t
t = 2.75s


11 = 0.5 ( 16 + 0)t
t = 1.375s
 
  • #14
Those answers are incorrect and are not expressed in terms of "a".
 
  • #15
I don't see any other way to relate acceleration and time
 
  • #16
v = at
 
  • #17
Thats what I used In post #11
 
  • #18
Ab17 said:
Thats what I used In post #11
Post 11 claims that 8 = 16.
 
  • #19
I will get the same thing if I use V = at
 
  • #20
Ab17 said:
I will get the same thing if I use V = at
Let us review...

jbriggs444 said:
How much time elapses while the object slows down from -8 m/s to 0 m/s?
How much time elapses while the object speeds up from 0 m/s to 16 m/s?
Express both answers in terms of a.

The answer for the first question is "the elapsed time is 8/a"
The answer for the second question is "the elapsed time is 16/a"

If the acceleration is the same for both phases, what does that say about the elapsed time for both phases?
 
  • #21
Time has to be different
 
  • #22
Ab17 said:
Time has to be different
Yes. The time has to be different. And the ratio between the two times is what?
 
  • #23
8/a and 16/a
 
  • #24
That would be 2 to 1. It takes twice as long accelerating to 16 m/s as it does decelerating from 8 m/s.

Now, If the ratio of the times is 2 to 1, what is the ratio of the distances traveled during the two phases?
 
  • #25
2 to 1?
 
  • #26
Ab17 said:
2 to 1?
If acceleration is fixed and you double the time, what happens to the distance traversed?
 
  • #27
X= 1/2axt^2
 
  • #28
So what is the ratio of the two distances?
 

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