Rectilinear Kinematics
Average Velocity, Average Speed & Acceleration from a Position Function
A particle moves along a straight line with its position defined by
s = (t² − 6t + 5) m, where t is in seconds.
Determine the average velocity, the average speed, and the acceleration
of the particle over the interval 0 to t = 6 s.
What We Know
- Position function: s(t) = t² − 6t + 5 (m)
- Time interval: t = 0 to t = 6 s
- Find: average velocity over [0, 6 s]
- Find: average speed over [0, 6 s]
- Find: acceleration at t = 6 s
Strategy:
(1) Differentiate s(t) once to get v(t); set v = 0 to find any turning points.
(2) Evaluate s at t = 0, the turning point, and t = 6 s.
(3) Average velocity = net displacement / time (signed).
(4) Average speed = total distance / time (always positive — requires segment-by-segment counting).
(5) Differentiate v(t) to get a(t) — since v(t) is linear, acceleration is constant.
Part A — Velocity Function & Turning Point
Step 1
Differentiate s(t) to Get v(t) and Find the Turning Point
Velocity is the time derivative of position:
\[ v(t) = \frac{ds}{dt} = 2t – 6 \quad \text{m/s} \]
Check the initial velocity at t = 0 to understand the starting motion:
\[ v(0) = 2(0) – 6 = -6 \ \text{m/s} \]
The particle starts moving in the negative direction. It reverses when v = 0:
\[ 2t – 6 = 0 \quad \Rightarrow \quad t = 3 \ \text{s} \]
\[ \text{Turning point at } t = 3 \ \text{s} \]
Physical note: For 0 < t < 3 s, v < 0 (particle moves left). For 3 < t < 6 s, v > 0 (particle moves right). This single reversal means the interval [0, 6 s] has two motion segments that must be counted separately for distance.
Part B — Position at Key Instants
Step 2
Evaluate s(t) at t = 0, 3, and 6 s
Substitute each critical time into the position function:
\[ s(0) = (0)^2 – 6(0) + 5 = 5 \ \text{m} \]
\[ s(3) = (3)^2 – 6(3) + 5 = 9 – 18 + 5 = -4 \ \text{m} \]
\[ s(6) = (6)^2 – 6(6) + 5 = 36 – 36 + 5 = 5 \ \text{m} \]
| Time (s) | Position (m) | Event |
|---|---|---|
| 0 | +5 | Start |
| 3 | −4 | Turning point (leftmost) |
| 6 | +5 | End |
Interesting observation: s(0) = s(6) = 5 m — the particle returns to exactly the same position it started from! This means the net displacement is zero, so the average velocity will be zero. Yet the particle clearly traveled a real distance, making average speed non-zero.
Part C — Average Velocity
Step 3
Compute Average Velocity
Average velocity is the net displacement divided by the elapsed time:
\[ \bar{v} = \frac{s(6) – s(0)}{\Delta t} = \frac{5 – 5}{6 – 0} = \frac{0}{6} \]
\[ \bar{v} = 0 \ \text{m/s} \]
Why zero? Average velocity only measures the straight-line change from start to finish. Since the particle ends up at the same position it began, its net displacement is zero — regardless of how far it actually traveled in between. This is a perfect illustration of why displacement and distance must never be confused.
Part D — Average Speed
Step 4
Compute Total Distance and Average Speed
Total distance requires summing the absolute length of each segment:
\[ d_\text{total} = |s(3) – s(0)| + |s(6) – s(3)| \]
\[ = |-4 – 5| + |5 – (-4)| = 9 + 9 \]
\[ d_\text{total} = 18 \ \text{m} \]
Average speed is total distance divided by elapsed time:
\[ \bar{v}_\text{speed} = \frac{d_\text{total}}{\Delta t} = \frac{18 \ \text{m}}{6 \ \text{s}} \]
\[ \bar{v}_\text{speed} = 3 \ \text{m/s} \]
Segment breakdown:
Segment 1 (t = 0 to 3 s): particle moves left from s = 5 m to s = −4 m → 9 m traveled.
Segment 2 (t = 3 to 6 s): particle moves right from s = −4 m back to s = 5 m → 9 m traveled.
Both segments are equal in length — a symmetric journey. Total = 18 m.
Part E — Acceleration at t = 6 s
Step 5
Differentiate v(t) to Find Acceleration
Acceleration is the time derivative of velocity:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(2t – 6) = 2 \ \text{m/s}^2 \]
\[ a = 2 \ \text{m/s}^2 \quad \text{(constant — independent of } t\text{)} \]
Physical note: Because s(t) is a quadratic (degree 2) in time, its second derivative is a non-zero constant — meaning the particle undergoes uniform acceleration throughout the entire 6-second interval. The value a = +2 m/s² means the acceleration is always directed to the right, which is why the initial leftward motion eventually reverses: the constant rightward acceleration steadily erodes the negative velocity until it crosses zero at t = 3 s.
Final Answers
AVERAGE VELOCITY (0 to 6 s)
v̄ = 0 m/s (net displacement = 0)
AVERAGE SPEED (0 to 6 s)
v̄speed = 3 m/s (total distance = 18 m)
ACCELERATION AT t = 6 s
a = 2 m/s² (constant throughout)