For motion along a one-dimensional path (straight or curved), define
\(s =\) displacement from the origin along the path
\(v =\) velocity along the path
\(a =\) acceleration along the path
These quantities are related by the two basic differential relations:
\[ds = v \; dt\] \[dv = a \; dt\]
If you use them to eliminate \(dt\), you also have
\[a \; ds = v \; dv\]
With these three differential relations one can solve many kinematics problems, as shown in the table below. Examples of each are given afterward.
| Given | Find | kinematic equation | |
|---|---|---|---|
| 1 | \(s(t)\) | \(v(t)\) | \(\displaystyle v(t) = \frac{ds}{dt}\) |
| 2 | \(v(t)\) | \(a(t)\) | \(\displaystyle a(t) = \frac{da}{dt}\) |
| 3 | \(v(s)\) | \(a(s)\) | \(\displaystyle a(s) = v \frac{dv}{ds}\) |
| 4 | \(v(t)\) | \(s(t)\) | \(\displaystyle \int_{t_0}^{t} v(t)\; dt = \int_{s_0}^{s} ds = s(t) - s_0\) |
| 5 | \(a(t)\) | \(v(t)\) | \(\displaystyle \int_{t_0}^{t} a(t)\; dt = \int_{v_0}^{v} dv = v(t) - v_0\) |
| 6 | \(a(s)\) | \(v(s)\) | \(\displaystyle \int_{s_0}^{s} a(s)\; ds = \int_{v_0}^{v}v\; dv = \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2\) |
| 7 | \(v(s)\) | \(s(t)\) | \(\displaystyle dt = \frac{ds}{v} \; \to \; \int_{t_0}^{t} dt = t-t_0 = \int_{s_0}^{s} \frac{ds}{v(s)}\) |
| 8 | \(a(v)\) | \(v(t)\) | \(\displaystyle dt = \frac{dv}{a} \; \to \; \int_{t_0}^{t} dt = t - t_0 = \int_{v_0}^{v} \frac{dv}{a(v)}\) |
Notes
The position of the particle is given by \(s = (2t^2 - 8t + 6)\) m, where \(t\) is in seconds. Determine the time when the velocity of the particle is zero.
Velocity is \(\displaystyle v(t) = \frac{ds}{dt} = 4t -8\) m/s.
Set velocity to zero and solve for time: \(4t -8 = 0\), so
\[t = 2 \; \rm{s}\]
A particle travels along a straight line with a speed \(v = (0.5 t^3 - 8t)\) m/s, where \(t\) is in seconds. Determine the acceleration of the particle when \(t=2\) s.
Acceleration is \(\displaystyle a(t) = \frac{dv}{dt} = 1.5 t^2 -8\).
Set time to \(t=2\) s and solve for \(a\):
\[a = -2 \; \rm{m/s}^2\]
A particle travels along a straight line with a velocity of \(v = (20 - 0.05s^2)\) m/s, where \(s\) is in meters. Determine the acceleration of the particle at \(s=15\) m.
Acceleration is found by \(\displaystyle a(s) = v \frac{dv}{ds}\).
We need the derivative \(\displaystyle \frac{d}{ds} (20 - 0.05s^2) = -0.1s\).
Then \(\displaystyle a(s) = v \frac{dv}{ds} = (20 - 0.05s^2)(-0.1s) = -2s +0.005s^3\).
So at \(s=15\) m,
\[a = -13.1 \rm{m/s}^2\]
A particle travels along a straight line with a velocity of \(v = (4t - 3t^2)\) m/s, where \(t\) is in seconds. Determine the position of the particle when \(t=4\) s, given that \(s = 0\) when \(t = 0\).
Displacement is given by \(\displaystyle \int_{t_0}^{t} v(t)\; dt = s(t) - s_0\).
Initial conditions are \(t_0 = 0\) and \(s_0 = 0\). So
\[s(t) = \int_{0}^{t} (4t - 3t^2) \; dt = ½ \cdot 4 t^2 - ⅓ \cdot 3 t^3 = 2 t^2 - t^3 \]
At \(t=4\) s,
\[s = -32 \; \rm{m}\]
Starting from rest, a particle moving in a straight line has an acceleration of \(a = (2t - 6)\) m/s², where \(t\) is in seconds. What is the particle’s velocity when \(t=6\) s?
Velocity is given by \(\displaystyle \int_{t_0}^{t} a(t)\; dt = \int_{v_0}^{v} dv = v(t) - v_0\).
Initial conditions are \(t_0 = 0\) and \(v_0 = 0\) (since we are told it starts from rest). So
\[v(t) = \int_{0}^{t} (2t - 6) \; dt = ½ \cdot 2 t^2 - 6 t = t^2 - 6t \]
At \(t=6\) s,
\[v = 0 \; \rm{m/s}\]
A particle travels along a straight line with an acceleration of \(a = (10 - 0.2s)\) m/s², where \(s\) is measured in meters. Determine the velocity of the particle when \(s=10\) m if \(v=5\) m/s at \(s=0\).
Acceleration is given by \(\displaystyle \int_{s_0}^{s} a(s)\; ds = \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2\)
Initial conditions are \(v_0 = 5\) m/s and \(s_0 = 0\). So
\[ \tfrac{1}{2} v^2(s) - \tfrac{1}{2} \cdot 25 = \int_{0}^{s} (10 - 0.2s) \; ds = 10s - ½ \cdot 0.2 s^2 = 10s - 0.1 s^2 \]
Solve this for \(v(s)\):
\[v(s) = \sqrt{ 20s - 0.2 s^2 + 25 }\]
At \(s=10\) m,
\[v = 14.3 \; \rm{m/s}\]
The velocity of a particle traveling along a straight line is \(v = v_0 - ks\), where \(k\) is constant. If \(s=0\) when \(t=0\), determine the position of the particle as a function of time.
Displacement \(s(t)\) is found by \(\displaystyle t-t_0 = \int_{s_0}^{s} \frac{ds}{v(s)}\).
Initial conditions are \(s=0\) at time \(t=0\). So
\[t = \int_{0}^{s} \frac{ds}{(v_0 - ks)} = \left[ -\frac{1}{k} \ln (v_0 - ks) \right]_0^s = -\frac{1}{k} \ln \frac{v_0-ks}{v_0}\]
Solve by exponentiating both sides: \(\displaystyle \frac{v_0-ks}{v_0} = e^{-kt}\), so
\[s(t) = \frac{v_0}{k} \left(1 - e^{-kt} \right)\]
A particle is moving with a velocity of \(v_0\) when \(s=0\) and \(t=0\). If it is subjected to a deceleration of \(a = -kv^3\), where \(k\) is a constant, determine its velocity as a function of time.
Velocity is found by \(\displaystyle t - t_0 = \int_{v_0}^{v} \frac{dv}{a(v)}\).
Initial conditions are \(v=v_0\) at time \(t_0=0\) and position \(s=0\). So
\[t = \int_{v_0}^{v} \frac{ds}{-kv^3} = \left[ \frac{1}{2k v^2} \right]_{v_0}^v = \frac{1}{2k} \left( \frac{1}{v^2} - \frac{1}{v_0^2} \right)\]
Solving for \(v\): \(\frac{1}{v_0^2} + 2kt = \frac{1}{v^2}\), so
\[v(t) = \sqrt{ \frac{v_0^2}{ 1+ 2kt v_0^2 } }\]