Rectilinear Kinematics
Deceleration, Distance & Average Speed from a Velocity Function
A particle moves along a straight line with a velocity given by
v = (6t − 3t²) m/s, where t is in seconds.
The particle starts at the origin (s = 0) when t = 0.
Find the deceleration and position of the particle at t = 3 s,
the total distance it has traveled over the 3-second interval,
and its average speed over that same period.
What We Know
- Velocity function: v(t) = (6t − 3t²) m/s
- Initial condition: s = 0 m at t = 0 s
- Time interval of interest: 0 to 3 s
- Find: deceleration at t = 3 s
- Find: position at t = 3 s
- Find: total distance traveled over [0, 3 s]
- Find: average speed over [0, 3 s]
Strategy: Four quantities, three tools.
(1) Differentiate v(t) to get a(t), then evaluate at t = 3 s.
(2) Integrate v(t) with the initial condition to get s(t), evaluate at t = 3 s.
(3) Locate the turning point (v = 0) inside [0, 3 s] to correctly compute total distance — displacement and distance differ when the particle reverses.
(4) Average speed = total distance / total time.
Part A — Deceleration at t = 3 s
Step 1
Differentiate v(t) to Find a(t)
Acceleration is the time derivative of velocity. Differentiating the given velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t – 3t^2) = 6 – 6t \quad \text{m/s}^2 \]
Substituting t = 3 s:
\[ a\big|_{t=3} = 6 – 6(3) = 6 – 18 \]
\[ a = -12 \ \text{m/s}^2 \quad \Rightarrow \quad \text{Deceleration} = 12 \ \text{m/s}^2 \]
Physical note: The negative sign confirms the acceleration opposes the positive direction. “Deceleration” refers to the magnitude — 12 m/s². Notice that a(t) = 6 − 6t = 0 at t = 1 s; before t = 1 s the particle accelerates, and after t = 1 s it decelerates continuously — the deceleration grows linearly with time.
Part B — Position at t = 3 s
Step 2
Integrate v(t) to Obtain the Position Function s(t)
Since v = ds/dt, we separate variables and integrate with the initial condition s = 0 at t = 0:
\[ \int_{0}^{s} ds = \int_{0}^{t} (6t – 3t^2)\, dt \]
\[ s = \Big[3t^2 – t^3\Big]_{0}^{t} \]
\[ s(t) = 3t^2 – t^3 \quad \text{m} \]
Evaluating at t = 3 s:
\[ s\big|_{t=3} = 3(3)^2 – (3)^3 = 27 – 27 \]
\[ s = 0 \ \text{m} \]
Physical interpretation: The particle returns exactly to its starting point at t = 3 s! This means the net displacement over the full 3-second interval is zero — the particle traveled out and came all the way back. However, the total distance traveled is definitely not zero, as we will show in Part C.
Part C — Total Distance & Average Speed
Step 3
Find the Turning Point (v = 0)
The particle changes direction wherever velocity equals zero. Setting v(t) = 0:
\[ 6t – 3t^2 = 0 \quad \Rightarrow \quad 3t(2 – t) = 0 \]
\[ t = 0 \quad \text{or} \quad t = 2 \ \text{s} \]
The relevant turning point within (0, 3 s] is at t = 2 s. Find the position there:
\[ s(2) = 3(2)^2 – (2)^3 = 12 – 8 = 4 \ \text{m} \]
Motion summary: For 0 < t < 2 s, v > 0 (particle moves in positive direction). For 2 < t < 3 s, v < 0 (particle reverses and moves in the negative direction). The farthest point reached is s = 4 m at t = 2 s.
Step 4
Compute Total Distance and Average Speed
With the three key positions — s(0) = 0 m, s(2) = 4 m, s(3) = 0 m — the total path length is:
\[ d_\text{total} = |s(2) – s(0)| + |s(3) – s(2)| \]
\[ = |4 – 0| + |0 – 4| = 4 + 4 \]
\[ d_\text{total} = 8 \ \text{m} \]
Average speed is total distance divided by total elapsed time:
\[ \bar{v} = \frac{d_\text{total}}{t} = \frac{8 \ \text{m}}{3 \ \text{s}} \]
\[ \bar{v} = 2.67 \ \text{m/s} \]
Displacement vs. Distance — key distinction: Net displacement = s(3) − s(0) = 0 − 0 = 0 m. Total distance = 8 m. Average velocity (based on displacement) = 0 m/s. Average speed (based on distance) = 2.67 m/s. These differ because the particle reversed direction — a critical distinction in kinematics.
Final Answers
DECELERATION AT t = 3 s
|a| = 12 m/s²
POSITION AT t = 3 s
s = 0 m (back at the origin)
TOTAL DISTANCE TRAVELED (0 to 3 s)
d = 8 m
AVERAGE SPEED
v̄ = 8/3 ≈ 2.67 m/s