Kinematics

Kinematics

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.

QuantityPositiveZeroNegative
\(s\)Positive side of originAt the originNegative side of origin
\(v\)Moves in positive directionInstantaneously at restMoves in negative direction
\(a\)Velocity increasingVelocity stationaryVelocity 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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Answer:\(\displaystyle v=\frac{\mathrm{d}s}{\mathrm{d}t}\), \(\displaystyle a=\frac{\mathrm{d}v}{\mathrm{d}t}\), \(\displaystyle s=\int v\,\mathrm{d}t+C_s\), and \(\displaystyle v=\int a\,\mathrm{d}t+C_v\). At rest, \(v=0\). At a change of direction, \(v\) changes sign through \(0\). At an interior extremum of differentiable velocity, \(a=0\), with a sign change confirming the extremum.

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