Initial Acceleration Homework: Find V0, a, and x0

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Homework Help Overview

The discussion revolves around a kinematics problem involving a truck's motion along a straight line, represented by a quadratic position function. Participants are tasked with determining the initial velocity, acceleration, and position of the truck based on the provided equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the interpretation of the position function and its components, questioning the values of initial velocity and acceleration. Some participants attempt to derive acceleration from the position equation, while others explore the implications of time on the truck's motion.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's reasoning. Some have offered guidance on how to approach the problem, while others are exploring different interpretations of the equations involved. There is a recognition of the need to clarify the starting position and the implications of the truck's motion over time.

Contextual Notes

Participants are navigating constraints such as the specific form of the position equation and the implications of initial conditions. There is an ongoing examination of the units involved and how they relate to the physical scenario described.

  • #31
here comes another one, I believe I know the answer just making sure, as if I get this wrong it is automatically marked as wrong.

Prior to this question, I was asked what was the velocity (magnitude) of the truck at t = 6.20s,

I used:

Velocity = Initial Velocity + Acceleration (Time)
I managed to get that right :cool:

Now I need to know the velocity (direction) of the truck at t = 6.20s.

The two options I got is +x axis and -x axis.

The first thing that comes to my mind is that it is the +x axis, as time is increasing, hence +x axis.

Let me know if this is the correct logic to see this statement, or if I am wrong.

Please and thank you.
 
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  • #32
You're correct with your answer (+x axis), but I'm not sure about the logic...

You calculated v = v0 + at. If this number came out positive, then the direction is along the +x axis. If it came out negative then the direction is along the -x axis.
 
  • #33
learningphysics said:
You're correct with your answer (+x axis), but I'm not sure about the logic...

You calculated v = v0 + at. If this number came out positive, then the direction is along the +x axis. If it came out negative then the direction is along the -x axis.

Ok, so regardless of time, it will be based on the equation mentioned. Thanks for the clarification.
 
  • #34
ok, here's another.

As we know we solved the at what time is the truck first time a distance of 10.0 cm from its starting point.

It asked me for the second time. I used the second answer I got from the quadratic equation which is 25.8s.

Now its asking me for the third. The quadradic only gave me two answers.

So I did the following: 25.8-6.20 = 19.6 difference...so 25.8+19.6 = 45.4.

Would it not follow the same trend is all variables remained the same?
 
  • #35
Heat said:
Now its asking me for the third. The quadradic only gave me two answers.
Now it's time to use what learningphysics mentioned earlier: 10 cm from the start could be 50 +10 or 50 -10 cm.
 
  • #36
I would like to thanks Doc Al, learningphysics and G01for their kind contribution.:smile:
 

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