Motion along a straight line (Finding acceleration).

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SUMMARY

The discussion focuses on calculating the acceleration of a red car that starts from rest while moving towards a green car traveling at a constant speed of -20 m/s. The initial positions are given as xg0 = 270 m for the green car and xr0 = -35 m for the red car, with the two cars intersecting at t = 12 seconds. The user initially calculated the position using the formula X = Xo + (V - Vo/2)t, yielding an incorrect distance of 150 m, which led to an erroneous acceleration calculation of a = 1.68 m/s² instead of the correct value of a = 0.9 m/s². The error stemmed from misinterpretation of the distance traveled by the green car.

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  • Understanding of kinematic equations for linear motion
  • Familiarity with concepts of initial velocity and constant speed
  • Ability to interpret motion graphs
  • Knowledge of basic algebra for solving equations
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  • Review kinematic equations, specifically X = Xo + Vot + 1/2at²
  • Study the concept of relative motion in physics
  • Learn how to analyze motion graphs for accuracy
  • Practice problems involving acceleration and constant speed scenarios
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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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