Initial Acceleration Homework: Find V0, a, and x0

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SUMMARY

The discussion revolves around solving a physics homework problem involving a truck's motion along a straight line, described by the equation x(t) = 50.0 cm + (2.00 cm/s)t - (0.0625 cm/s²)t². The initial velocity is confirmed as 2.00 cm/s, and the initial position is 50.0 cm. The initial acceleration is determined to be -0.125 cm/s². The participants also explore the time taken for the truck to return to its starting point, concluding it takes 32 seconds, and discuss the truck's distance from its starting point at various times.

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  • #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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