Learning Objectives
Calculate velocity, speed, and acceleration from a position function.
Distinguish displacement from distance and velocity from speed.
Use a dot product to decide whether speed is increasing or decreasing.
Prerequisites: vector derivatives, magnitudes, and dot products. In the examples, position is in metres and time is in seconds unless stated otherwise.
1. Position, Velocity, and Acceleration
If r(t) is position, velocity is v(t) = r′(t), measured in metres per second. Acceleration is a(t) = v′(t) = r″(t), measured in metres per second squared. Both are vectors and contain directional information.
Speed is the scalar magnitude |v(t)|. It is nonnegative and does not record direction. An object can have constant speed while its velocity changes because it turns.
Worked Example: Motion on a Parabola
Let r(t) = ⟨t, t², 0⟩. Then v(t) = ⟨1, 2t, 0⟩ and a(t) = ⟨0, 2, 0⟩.
At t = 1, velocity is ⟨1, 2, 0⟩ m/s and speed is √5 m/s. Acceleration is ⟨0, 2, 0⟩ m/s².
The speed formula is √(1 + 4t²). Its units and scalar form differ from the velocity vector.
2. Acceleration Can Change Direction without Changing Speed
Worked Example: Uniform Circular Motion
Let r(t) = ⟨3 cos t, 3 sin t, 0⟩. Then v(t) = ⟨−3 sin t, 3 cos t, 0⟩.
Speed is √(9 sin²t + 9 cos²t) = 3, so it is constant. Acceleration is a(t) = ⟨−3 cos t, −3 sin t, 0⟩, which is nonzero.
The acceleration points toward the circle's center. It changes the direction of velocity rather than its magnitude.
3. A Dot Product Tests Change in Speed
At times when v(t) ≠ 0, differentiating |v|² = v · v gives d|v|/dt = (v · a)/|v|. Thus a positive v · a means speed is increasing, a negative value means speed is decreasing, and zero means the instantaneous rate of change of speed is zero.
Example: Speeding Up or Slowing Down
For r(t) = ⟨t, t², 0⟩, the dot product is ⟨1, 2t, 0⟩ · ⟨0, 2, 0⟩ = 4t.
The object speeds up for t > 0 and slows down for t < 0. At t = 0, the speed has derivative zero, even though the acceleration remains nonzero.
4. Displacement and Distance Measure Different Things
Displacement over [a, b] is r(b) − r(a), which also equals the vector integral of velocity. Total distance is the scalar integral of speed. A return trip may have zero displacement but positive distance.
Worked Example: A Full Revolution
For the circle of radius 3, travel from t = 0 to t = 2π. The starting and ending positions are both ⟨3, 0, 0⟩, so displacement is the zero vector.
Speed is constantly 3. The distance is the integral of 3 from 0 to 2π, giving 6π metres, equal to the circle's circumference.
Common Mistakes
Zero displacement does not mean no motion occurred.
Constant speed does not imply zero acceleration.
The speed-derivative formula divides by |v|, so it cannot be used unchanged when v = 0.
5. Practice
1. For r(t) = ⟨2t, t², 1⟩, find velocity, acceleration, and speed at t = 2.
2. If v = ⟨2, −1, 0⟩ and a = ⟨−3, 1, 0⟩ at an instant, is speed increasing or decreasing?
3. A particle has constant velocity ⟨3, 4, 0⟩ for 2 seconds. Find its displacement and distance traveled.
Worked Solutions
1. v(t) = ⟨2, 2t, 0⟩ and a(t) = ⟨0, 2, 0⟩. At t = 2, velocity is ⟨2, 4, 0⟩, acceleration is ⟨0, 2, 0⟩, and speed is √20 = 2√5.
2. v · a = −6 − 1 = −7, so speed is decreasing. The nonzero velocity makes the sign test applicable.
3. Displacement is 2⟨3, 4, 0⟩ = ⟨6, 8, 0⟩ metres. Speed is 5 m/s, so distance is 10 metres.
6. Summary
- Velocity and acceleration are vectors; speed is a magnitude.
- Use v · a to test speed change when velocity is nonzero.
- Integrate velocity for displacement and speed for distance.
- Check units and distinguish a changing direction from a changing speed.