Choose a positive direction. Displacement gives position relative to a fixed origin; velocity gives direction and speed.
\(v=\dfrac{\mathrm ds}{\mathrm dt}\qquad a=\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{\mathrm d^2s}{\mathrm dt^2}\)
\(v=\int a\,\mathrm dt\qquad s=\int v\,\mathrm dt\)
Find each integration constant from the given initial conditions. Do not assume the particle starts at the origin or from rest.
| Quantity | Positive | Zero | Negative |
|---|---|---|---|
| \(s\) | Positive side of origin | At the origin | Negative side of origin |
| \(v\) | Moves in positive direction | Instantaneously at rest | Moves in negative direction |
| \(a\) | Velocity increasing | Velocity stationary | Velocity decreasing |
Speed and direction
\(\text{Speed}=|v|\qquad va>0:\ \text{speeding up}\qquad va<0:\ \text{slowing down}\)
A direction change requires a change in the sign of velocity. Zero velocity alone is insufficient; zero acceleration alone does not establish constant velocity.
At an interior extremum of differentiable velocity, a = 0; test its nature.
Example
A particle has displacement \(s=t^3-9t^2+24t\) m from \(O\) at time \(t\) s. Then \(v=3t^2-18t+24=3(t-2)(t-4)\), so it is at rest at \(t=2\) and \(t=4\), changing direction each time.
\(s(0)=0,\quad s(2)=20,\quad s(4)=16,\quad s(5)=20\)
Distance in the first \(5\) s \(=20+4+4=28\) m, although the displacement at \(t=5\) is only \(20\) m.
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