Distance and displacement confusion

AI Thread Summary
The discussion centers on the confusion between distance and displacement in a physics problem involving two objects with constant acceleration. The first object has an initial velocity of -8.00 m/s and a final velocity of 16.0 m/s, with a displacement of 20.0 m, while the second object travels a total distance of 22.0 m. Participants clarify that the acceleration must be constant, emphasizing that the path cannot be split arbitrarily and that the time taken for each phase of motion must be expressed in terms of acceleration. They conclude that the time ratios for deceleration and acceleration phases are 2 to 1, which affects the distance traveled during each phase. The conversation highlights the importance of correctly applying kinematic equations and understanding the relationship between time, acceleration, and distance.
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 accleration 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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