Rectilinear Kinematics
Acceleration, Displacement & Distance from a Velocity Function
A particle travels along a straight line with a velocity described by
v = (12 − 3t²) m/s, where t is in seconds.
At t = 1 s, the particle is located 10 m to the left of the origin.
Determine: (1) the acceleration at t = 4 s,
(2) the displacement from t = 0 to t = 10 s,
and (3) the total distance traveled during this time period.
What We Know
- Velocity function: v(t) = (12 − 3t²) m/s
- Position at t = 1 s: s = −10 m (10 m to the left ←)
- Find: acceleration at t = 4 s
- Find: displacement from t = 0 to t = 10 s
- Find: total distance traveled from t = 0 to t = 10 s
Strategy overview: Three separate tasks require three tools.
(1) Acceleration = dv/dt — differentiate the velocity function.
(2) Position = integrate v dt with the given initial condition to get s(t), then compute Δs = s(10) − s(0).
(3) Distance requires locating any turning points (where v = 0) within [0, 10 s] and summing absolute segment lengths — displacement and distance differ whenever the particle reverses direction.
Part A — Acceleration at t = 4 s
Step 1
Differentiate v(t) to Get a(t)
Acceleration is the time derivative of velocity. Differentiating the given velocity function with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt}\left(12 – 3t^2\right) = -6t \quad \text{m/s}^2 \]
Substituting t = 4 s:
\[ a\big|_{t=4} = -6(4) \]
\[ a = -24 \ \text{m/s}^2 \]
Physical note: The negative sign means the acceleration acts in the negative direction (leftward). Because the acceleration a = −6t grows in magnitude over time, the particle decelerates more and more aggressively as time progresses. At t = 4 s it is already decelerating at 24 m/s².
Part B — Displacement from t = 0 to t = 10 s
Step 2
Integrate v(t) to Obtain the Position Function s(t)
Since v = ds/dt, separating variables and integrating with the known condition s = −10 m at t = 1 s:
\[ \int_{-10}^{s} ds = \int_{1}^{t} (12 – 3t^2)\, dt \]
\[ s – (-10) = \Big[12t – t^3\Big]_{1}^{t} \]
\[ s + 10 = \left(12t – t^3\right) – \left(12(1) – (1)^3\right) \]
\[ s + 10 = 12t – t^3 – 12 + 1 \]
\[ s(t) = -t^3 + 12t – 21 \quad \text{m} \]
Verification: At t = 1 s: s = −1 + 12 − 21 = −10 m ✓ — matches the given initial condition perfectly.
Step 3
Evaluate s(t) at t = 0 and t = 10 s
At t = 0 s:
\[ s_0 = -(0)^3 + 12(0) – 21 = -21 \ \text{m} \]
At t = 10 s:
\[ s_{10} = -(10)^3 + 12(10) – 21 = -1000 + 120 – 21 = -901 \ \text{m} \]
Displacement = final position − initial position:
\[ \Delta s = s_{10} – s_0 = -901 – (-21) \]
\[ \Delta s = -880 \ \text{m} \]
Physical interpretation: The negative displacement means the particle ends up 880 m to the left of where it started at t = 0. Displacement only cares about start and end positions — not the path taken in between.
Part C — Total Distance Traveled
Step 4
Locate the Turning Point (where v = 0)
Distance and displacement differ whenever the particle changes direction. A direction reversal occurs where the velocity equals zero. Setting v(t) = 0:
\[ 12 – 3t^2 = 0 \quad \Rightarrow \quad t^2 = 4 \quad \Rightarrow \quad t = 2 \ \text{s} \]
The only turning point within the interval [0, 10 s] is at t = 2 s. Compute the position there:
\[ s_2 = -(2)^3 + 12(2) – 21 = -8 + 24 – 21 = -5 \ \text{m} \]
Physical note: For t < 2 s, v > 0 (particle moves right). For t > 2 s, v < 0 (particle moves left). The turning point at t = 2 s is the rightmost position the particle reaches during the entire interval.
Step 5
Sum the Absolute Segment Lengths
With the turning point identified, the total distance is the sum of the absolute lengths of each motion segment:
\[ d_{\text{total}} = |s_2 – s_0| + |s_{10} – s_2| \]
\[ = |{-5} – ({-21})| + |{-901} – ({-5})| \]
\[ = |{+16}| + |{-896}| \]
\[ = 16 \ \text{m} + 896 \ \text{m} \]
\[ d_{\text{total}} = 912 \ \text{m} \]
Displacement vs. Distance — summary: The particle moves 16 m to the right (from −21 m to −5 m), then reverses and travels 896 m to the left (from −5 m to −901 m). Total path length = 912 m, but net displacement = −880 m. These differ because the particle reversed direction mid-journey.
Final Answers
ACCELERATION AT t = 4 s
a = −24 m/s² (directed to the left)
DISPLACEMENT (t = 0 to t = 10 s)
Δs = −880 m (880 m to the left)
TOTAL DISTANCE TRAVELED
d = 912 m