Displacement and Velocity Applications
Key Terms
- Velocity is the rate of change of displacement: v = ds/dt
- To find displacement from velocity: s = ∫ v dt + c; find c using s(t0) = s0
- Total distance ≠ displacement; distance = ∫ |v| dt (split the integral at zeros of v)
- Particle is at rest when v = 0; moving in positive direction when v > 0
- Speed = |v|; a particle can be slowing down even when moving in positive direction
v(t) = ds/dt a(t) = dv/dt = d²s/dt²
s(t) = ∫ v(t) dt + c v(t) = ∫ a(t) dt + c
Total Distance:
Split at all values of t where v(t) = 0.
Distance = ∫t1t2|v(t)| dt = sum of |sub-interval displacements|
Displacement = final position − initial position
(a) Find s(t). (b) When is the particle at rest? (c) Find displacement from t = 0 to t = 5.
(a) s(t) = ∫(2t − 6) dt = t² − 6t + c
s(0) = 0 − 0 + c = 3 ⇒ c = 3, so s(t) = t² − 6t + 3
(b) v(t) = 0: 2t − 6 = 0 ⇒ t = 3 s
(c) Displacement = s(5) − s(0) = (25 − 30 + 3) − 3 = −2 − 0 = −2 m
v = 0 at t = 3. Sign of v: negative on [0, 3], positive on [3, 5].
Distance = |s(3) − s(0)| + |s(5) − s(3)|
s(3) = 9 − 18 + 3 = −6
|−6 − 3| = 9 m |−2 − (−6)| = 4 m
Total distance = 9 + 4 = 13 m
The Language of Motion: Displacement, Velocity, and Speed
In one-dimensional motion, a particle moves along a line. Its position s(t) describes where it is at time t. Velocity v(t) is the rate of change of position: v = ds/dt. Speed is the magnitude of velocity: |v(t)|. These are related but distinct. A particle can have negative velocity (moving in the negative direction) while still having positive speed.
Displacement is the change in position: Δs = s(t2) − s(t1). It is a signed quantity — positive means net movement in the positive direction, negative means net movement in the negative direction. Total distance is the total length of the path travelled, regardless of direction, and is always non-negative.
Finding Position from Velocity
Since v(t) = ds/dt, we reverse this by integrating: s(t) = ∫ v(t) dt + c. The constant c is found from the initial condition — the position of the particle at t = 0 (or some other given time). This is identical to the process of finding f(x) from f′(x) studied in anti-differentiation.
Similarly, if acceleration a(t) = dv/dt is given, we integrate to get v(t) = ∫ a(t) dt + c, using an initial velocity to find c. Then integrate again to find s(t).
When Is the Particle at Rest?
The particle is at rest when its velocity is zero: v(t) = 0. Solve this equation to find the time(s) when the particle momentarily stops. These times are critical because they mark changes in direction — before t = 0, the particle moves one way; after, it may move the other way (if v changes sign).
To determine direction of motion: v > 0 means moving in the positive direction; v < 0 means moving in the negative direction. Sketch a sign diagram for v(t) to keep track of direction changes.
Computing Total Distance Travelled
The key challenge is that when the particle changes direction, the signed area under the velocity-time graph is no longer equal to total distance. The portion where v < 0 gives a negative contribution to displacement, but a positive contribution to distance.
The procedure is: (1) Find all zeros of v(t) in the interval [t1, t2]. (2) Split the interval at those zeros. (3) Evaluate |s(end) − s(start)| for each sub-interval. (4) Add all the absolute values.
Equivalently: Total distance = ∫t1t2 |v(t)| dt. Computing this integral directly requires splitting at sign changes of v.
Interpreting the Results
These applications connect integration to real physical meaning. The area under a v-t graph represents displacement. The area between the v-t graph and the t-axis (treating area below the axis as positive) represents total distance. This is why integration of velocity gives such powerful tools for understanding motion.
A typical exam question gives v(t), a time interval [0, T], and an initial position. You may be asked: (a) position at time T; (b) when the particle is at rest; (c) direction at a given time; (d) total distance over [0, T]; (e) distance from initial position at time T (which is simply |s(T) − s(0)|).
Mastery Practice
- A particle has velocity v(t) = 4t − 8 m/s. It starts at s = 0 when t = 0.
(a) Find s(t). (b) When is the particle at rest? (c) Find the displacement from t = 0 to t = 6. - A particle starts at position 5 m and has velocity v(t) = 3 − 6t m/s.
Find (a) position s(t), (b) position at t = 3. - A particle moves with velocity v(t) = t² − 4 m/s, t ≥ 0. Find when the particle is at rest.
- For v(t) = 4t − 8 from Q1, find the total distance travelled from t = 0 to t = 6.
- A particle has velocity v(t) = t² − 4t + 3 m/s and starts at s = 2 m.
(a) Find when the particle is at rest. (b) Find total distance from t = 0 to t = 4. - The velocity of a particle is v(t) = 6e−2t m/s, t ≥ 0, and its initial position is s(0) = 4 m.
Find s(t) and s(2). Give exact answers. - A particle starts at rest at the origin. Its acceleration is a(t) = 6t − 6 m/s².
(a) Find v(t). (b) Find s(t). (c) Find the total distance travelled from t = 0 to t = 3. - At t = 0 a particle is at position −3 m and has velocity 5 m/s. Its acceleration is a constant 2 m/s².
Find the time(s) when the particle is at the origin. - A particle moves along a line with velocity v(t) = t² − 5t + 4 m/s, t ∈ [0, 5]. At t = 0, s = 0.
(a) Find all times when the particle changes direction.
(b) Find the total distance travelled from t = 0 to t = 5. - A stone is thrown vertically upward from a point 2 m above the ground with initial velocity 15 m/s. Taking upward as positive, acceleration is −10 m/s².
(a) Find expressions for v(t) and s(t) where s is height above ground.
(b) Find the maximum height reached.
(c) Find the total distance travelled from t = 0 until the stone hits the ground.