anonymous
  • anonymous
Which of this statement is correct according to free Fall and kinematic equations?If its false explain why a. In general the value of V initial changes with time. b. In general the value of V final changes with time. c. In general the value of acceleration changes with time t. d. The signs of V initial, V final , acceleration and x final-X initial must be postive. e. In general the value of |x final-x initial| is the travel distance of the object up to time
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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theEric
  • theEric
Can you rule out any options?
theEric
  • theEric
Out of A, B, C, D, and E, do you see any that could not be right?
anonymous
  • anonymous
i fee like all is right except for e what do you think

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theEric
  • theEric
So, you're sure it is not E?
theEric
  • theEric
If you're not sure, we can talk about it. I think it would be most beneficial to help you to eliminate options, or approve of options. That's where the learning will happen, so you're better prepared for the future! And, you will have to prepare for the future if this is the beginning of the course, which I assume it is based on the material.
theEric
  • theEric
That, or this is a final...
anonymous
  • anonymous
no its the begining
anonymous
  • anonymous
i think abcd are right lookng at most formulas they all have to do with change in time but E is kind of debatable
theEric
  • theEric
Ah, but the question says that there is one correct statement that you have to pick!
anonymous
  • anonymous
It says circle each item below that is true . if an item is false explain y
anonymous
  • anonymous
so there could be several correct answerrs
theEric
  • theEric
Oh, right! Okay. So I'll tell you that not all of A, B, C, and D are true. Let's look at E. "In general the value of |x final-x initial| is the travel distance of the object up to time" Is that how it is worded?
anonymous
  • anonymous
yeah
theEric
  • theEric
Does it mean \(\left| s_\text{final}-x_\text{initial}\right|\) is the distance traveled upwards?
anonymous
  • anonymous
no ohh ok
theEric
  • theEric
\(\left| x_\text{final}-x_\text{initial}\right|\) I mean.
anonymous
  • anonymous
no it doesnt mean that its the distance travelled upwards
theEric
  • theEric
Ah, I think it means that it is the distance traveled within the time interval, then. \(x_f=x_i+v_i\ \Delta t+\frac{1}{2}a\ (\Delta t)^2\\\implies x_f-x_i=v_i\ \Delta t+\frac{1}{2}a\ (\Delta t)^2\) That is the sort of equation, probably. It is not true, depending on how you see the statement. \(\left| x_\text{final}-x_\text{initial}\right|\) is the magnitude of the change in position. That is actually displacement. If you throw a ball 10 meters up in the air, and it comes back to your hand, \(\left| x_\text{final}-x_\text{initial}\right|=0\). However, the distance traveled was \(10\ [m] +10\ [m]=20\ [m]\), so 20 meters. Thus, \(\left| x_\text{final}-x_\text{initial}\right|\) is not the same as the distance traveled in kinematic equations, even those with constant acceleration. In free fall , though, it is true. The ball is only going in one direction. I'm just not sure about the wording. There are tons of kinematic equations, but I guess we just mean free fall. So I'd say E is true.
theEric
  • theEric
I think there are three false statements, though.
theEric
  • theEric
\(v_i\) is constant, it is a fact and does not depend on time. Does acceleration depend on time? Look at a projectile. Should the acceleration be different at different times? It is always \(-9.8\ [m/s^2]\) right?
theEric
  • theEric
And, speaking of \(-9.8\ [m/s^2]\), is it always positive? :P Best of luck! I have to go!
theEric
  • theEric
I did give you information that will help you find the false statements and why they are false. Good luck!
anonymous
  • anonymous
thank you

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