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

- When integrating a differential relation, make sure limits on both
sides of the equation agree. For example, integrating \(v \; dt = ds\) gives \(\int_{t_0}^{t} v(t)\; dt = \int_{s_0}^{s}
ds\), which implies that
- at time \(t=t_0\) position is \(s=s_0\) (lower limits agree)
- at time \(t\) position is \(s=s(t)\) (upper limits agree)

- For one-dimensional motion along a curved path, the acceleration \(a\) above refers to only the tangential component of the acceleration (\(a_t\)).
- You may sometimes need more than one of these methods for a problem.
For example
- given \(s(t)\) find \(a(t)\) : use 1 then 2
- given \(a(s)\) find \(s(t)\) : use 6 then 7
- etc.

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

\[v(t) = \sqrt{ \frac{v_0^2}{ 1+ 2kt v_0^2 } }\]