Kinematics w/ Constant Acceleration

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Discussion Overview

The discussion revolves around a kinematics problem involving a particle moving with constant acceleration. Participants explore the relationships between displacement, velocity, and acceleration over time, particularly focusing on the conditions given in the problem statement.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the problem and attempts to derive equations for displacement and velocity, noting the conditions at specific times.
  • Another participant confirms that a zero velocity at t = 6 seconds indicates a zero slope on the s versus t plot at that point.
  • Equations for displacement and velocity are discussed, including the use of initial conditions to solve for unknowns.
  • Participants derive expressions for acceleration and velocity based on the given conditions, leading to numerical values for both.
  • There is a question about the reasoning behind solving for initial velocity first, which is explained as a trial-and-error approach to using the kinematics equations.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the problem and the derived values for velocity and acceleration. However, there is no explicit consensus on the correctness of the final numerical answers, as one participant asks for confirmation of their solution.

Contextual Notes

The discussion includes assumptions about the applicability of constant acceleration equations and the interpretation of the conditions provided in the problem. There may be limitations in the clarity of the initial conditions and the steps taken to derive the final results.

Who May Find This Useful

Students studying kinematics, particularly those working on problems involving constant acceleration and the relationships between displacement, velocity, and time.

aznkid310
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Homework Statement


A particle moves along the s-direction with constant acceleration. The displacement, measured from a convenient position, is 2m when t = 0 and is zero when t = 10s. If the velocity is momentarily zero when t = 6s, determine the acceleration a and the velocity v when t = 10s


Homework Equations



I tried finding plotting s vs. t and then trying to get an equation of the line, but that led me nowhere. Taking its derivative didnt help. I also looked at the formulas for constant acceleration but didnt know how to apply them. The fact that the velocity is momentarily zero at t = 6 means that the slope of my s vs. t plot is zero for that portion right?

The Attempt at a Solution



Equation of line: slope m = 0.5t
s = 0.5t + 2

Second attempt: s = s_0 + (v_0)t + 0.5at^2

s = 0, s_0 = 2
 
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aznkid310 wrote:[/color] "The fact that the velocity is momentarily zero at t = 6 means the slope of my s versus t plot is zero [at t = 6], right?"

Correct. v = ds/dt; therefore, if v(6) = 0, it means the instantaneous slope of the s versus t plot is zero at t = 6. But don't worry about plotting s versus t right now.

(1) Given (known).
s0 = 2.
s = 0 at t = 10.
v = 0 at t = 6.
Find a and v at t = 10.

(2) Relevant equations.
s = s0 + v0*t + 0.5*a*t^2.
v = v0 + a*t.

(3) Hint: Use the above relevant equations, and the above given conditions, to first solve simultaneously for v0.
 
1) s = 2 + v0*t + 0.5*a*t^2

2) v(6) = v0 + a(6) = 0
a = -v0/6

3) s = 2 + v0*t + 0.5*(-v0/6)*t^2

4) At t = 10s: 0 = 2 + 10(v0) + ([-100*v0]/12)

v0 = -1.2 m/s

5) v = v0 + a*t = -1.2 +a*t

6) At t = 6s: 0 = -1.2 + 6a

a = 0.2 m/s^2

7) v = -1.2 + 0.2t

8) At t = 10s: v = -1.2 +0.2*10

v = 0.8 m/s

Is this correct?
How did you know to solve for v0 first?
 
You did excellent work, aznkid310. That's correct.

How did I know to solve for v0 first? More or less, by trial and error. I just started plugging each given (known) condition into the various constant-acceleration kinematics formulas (experimenting, trying, investigating, retrying), to see what I would get and to see what is still unknown. I ended up with two acceptable-looking equations and two unknowns; so I knew I could at least go ahead and solve for those two unknowns, which I thought might help lead to a solution. Afterwards, computing v became straightforward.
 

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