Motion along a straight line (Finding acceleration).

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Homework Help Overview

The problem involves two cars moving toward each other, with one car having a constant speed and the other starting from rest. The objective is to determine the acceleration of the red car based on their positions and the time at which they meet.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find the position and acceleration of the red car, but questions arise regarding the formulas used and the interpretation of the graph.

Discussion Status

Participants are exploring the original poster's calculations and questioning the correctness of the formulas applied. There is a focus on clarifying the approach to finding the distance traveled by the green car and how it affects the overall interpretation of the problem.

Contextual Notes

There is a request for the figures associated with the problem, indicating that visual context may be necessary for a complete understanding. The discussion also highlights potential misinterpretations of the data provided in the problem statement.

Sarah Sposato
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Figure 2-24 shows a red car and a green car that move toward each other. Figure 2-25 is a graph of their motion, showing the positions xg0 = 270 m and xr0 = -35 m at time t = 0. The green car has a constant speed of -20 m/s and the red car begins from rest. What is the acceleration magnitude of the red car?
The graph of their motion shows the two cars intersect their position at t = 12 seconds.

I started out by interpreting the labels and information.
Xo (g) : 270m
Xo (r) : -35m
(Constant) V (g) : -20m/s
Vo (r) : 0 m/s.
t = 12 seconds.

1. I used the formula X = Xo + ( V - Vo/ 2 )t to get the position at which the cars meet.

X = 270 + (-20/2)12

X = 150m,

Then I used the formula X = Xo + volt + 1/2at^2 to get the acceleration of the red car.

150 = -35 + 0 (12) + 1/2a(12)^2.

This yielded a = 1.68 m/s*sThe answer is supposed to be a= 0.9m/s*s.Where did I go wrong? Was my interpretation of the data incorrect? Or just the way I plugged in the data?
 
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Can you post the two figures associated with this problem?
 
11668050_1142578102425170_817929921_n.jpg
 
Sarah Sposato said:
Figure 2-24 shows a red car and a green car that move toward each other. Figure 2-25 is a graph of their motion, showing the positions xg0 = 270 m and xr0 = -35 m at time t = 0. The green car has a constant speed of -20 m/s and the red car begins from rest. What is the acceleration magnitude of the red car?
The graph of their motion shows the two cars intersect their position at t = 12 seconds.

I started out by interpreting the labels and information.
Xo (g) : 270m
Xo (r) : -35m
(Constant) V (g) : -20m/s
Vo (r) : 0 m/s.
t = 12 seconds.

1. I used the formula X = Xo + ( V - Vo/ 2 )t to get the position at which the cars meet.
Why? The green car has a constant speed of -20 m/s, according to the problem statement.

What's the correct formula for finding distance traveled for an object traveling at constant speed?

X = 270 + (-20/2)12

X = 150m,
This distance X doesn't appear to be correct, just by looking at the graph which shows the motion of the cars versus time.

Then I used the formula X = Xo + volt + 1/2at^2 to get the acceleration of the red car.

150 = -35 + 0 (12) + 1/2a(12)^2.

This yielded a = 1.68 m/s*s

The answer is supposed to be a= 0.9m/s*s.

Where did I go wrong? Was my interpretation of the data incorrect? Or just the way I plugged in the data?
I think the second part of your work went wrong because you made the mistakes in the first part, as discussed above.
 

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