Rectilinear Kinematics
Velocity & Position by Integration — Variable Acceleration
A particle moves along a straight line with an acceleration given by
a = (2t − 1) m/s², where t is in seconds.
At t = 0, the particle is at s = 1 m with a velocity of v = 2 m/s.
Find the particle’s velocity and position at t = 6 s,
and determine the total distance traveled over this 6-second interval.
What We Know
- Acceleration function: a(t) = (2t − 1) m/s²
- Initial velocity: v₀ = 2 m/s at t = 0
- Initial position: s₀ = 1 m at t = 0
- Find: velocity at t = 6 s
- Find: position at t = 6 s
- Find: total distance traveled from t = 0 to t = 6 s
Strategy — integrate twice:
Since a(t) is a function of time (not constant), the simple kinematic formulas don’t apply.
Instead: (1) integrate a(t) with initial condition v₀ = 2 m/s to get v(t);
(2) integrate v(t) with initial condition s₀ = 1 m to get s(t);
(3) check whether v(t) = 0 has real solutions in [0, 6 s] to determine if a direction reversal occurs — this decides whether total distance equals displacement or must be computed separately.
Part A — Velocity at t = 6 s
Step 1
Integrate a(t) to Find v(t)
Starting from the definition a = dv/dt, separate variables and integrate with the initial condition v = 2 m/s at t = 0:
\[ \int_{2}^{v} dv = \int_{0}^{t} (2t – 1)\, dt \]
\[ \Big[v\Big]_{2}^{v} = \Big[t^2 – t\Big]_{0}^{t} \]
\[ v – 2 = t^2 – t \]
\[ v(t) = t^2 – t + 2 \quad \text{m/s} \qquad \cdots (1) \]
Substituting t = 6 s into equation (1):
\[ v(6) = (6)^2 – (6) + 2 = 36 – 6 + 2 \]
\[ v = 32 \ \text{m/s} \]
Physical note: The initial acceleration a(0) = 2(0) − 1 = −1 m/s² is negative, meaning the particle initially decelerates slightly. However, the acceleration becomes positive at t = 0.5 s and grows linearly from there, causing the velocity to climb steeply — reaching 32 m/s by t = 6 s.
Part B — Position at t = 6 s
Step 2
Integrate v(t) to Find s(t)
Using v = ds/dt, separate variables and integrate with the initial condition s = 1 m at t = 0:
\[ \int_{1}^{s} ds = \int_{0}^{t} (t^2 – t + 2)\, dt \]
\[ \Big[s\Big]_{1}^{s} = \left[\frac{t^3}{3} – \frac{t^2}{2} + 2t\right]_{0}^{t} \]
\[ s – 1 = \frac{t^3}{3} – \frac{t^2}{2} + 2t \]
\[ s(t) = \frac{t^3}{3} – \frac{t^2}{2} + 2t + 1 \quad \text{m} \qquad \cdots (2) \]
Substituting t = 6 s into equation (2):
\[ s(6) = \frac{(6)^3}{3} – \frac{(6)^2}{2} + 2(6) + 1 = 72 – 18 + 12 + 1 \]
\[ s = 67 \ \text{m} \]
Verification of initial condition: At t = 0: s = 0 − 0 + 0 + 1 = 1 m ✓ — the integration constant was correctly absorbed into the +1 term.
Part C — Total Distance Traveled
Step 3
Check for Turning Points — Does v(t) = 0 Have Real Solutions?
Before computing distance, we must determine whether the particle ever reverses direction in [0, 6 s]. Setting v(t) = 0:
\[ t^2 – t + 2 = 0 \]
Check the discriminant:
\[ \Delta = b^2 – 4ac = (-1)^2 – 4(1)(2) = 1 – 8 = -7 \]
\[ \Delta = -7 < 0 \quad \Rightarrow \quad \text{No real roots — velocity never reaches zero} \]
Physical meaning: The velocity function v(t) = t² − t + 2 is an upward-opening parabola with its minimum at t = 0.5 s, where v(0.5) = 0.25 − 0.5 + 2 = 1.75 m/s > 0. The particle is always moving in the positive direction — it never reverses. Therefore, total distance = displacement.
Step 4
Compute Total Distance
Since there is no direction reversal, total distance equals the magnitude of displacement:
\[ d_\text{total} = |s(6) – s(0)| = |67 – 1| \]
\[ d_\text{total} = 66 \ \text{m} \]
| Quantity | Expression | At t = 0 | At t = 6 s |
|---|---|---|---|
| Acceleration a | 2t − 1 m/s² | −1 m/s² | 11 m/s² |
| Velocity v | t² − t + 2 m/s | 2 m/s | 32 m/s |
| Position s | t³/3 − t²/2 + 2t + 1 m | 1 m | 67 m |
Key takeaway: This problem demonstrates why checking for turning points is always necessary — even when it seems obvious the particle moves in one direction. Here the discriminant test proves it rigorously: Δ < 0 means the velocity parabola never touches zero, so distance and displacement are identical. In contrast, problems 12-5, 12-6, and 12-7 all had turning points that required segment-by-segment counting.
Final Answers
VELOCITY AT t = 6 s
v = 32 m/s
POSITION AT t = 6 s
s = 67 m
TOTAL DISTANCE TRAVELED (0 to 6 s)
d = 66 m (no reversal — distance = displacement)