Kinematics w/ Constant Acceleration

[aznkid310]

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

 

[nvn]

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.

 

[aznkid310]

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?

 

[nvn]

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.