Rectilinear Kinematics
Position & Total Distance from a Cubic Position Function
The position of a particle along a straight line is described by
s = (1.5t³ − 13.5t² + 22.5t) ft,
where t is in seconds.
Find the position of the particle at t = 6 s,
and determine the total distance it travels over the 6-second interval.
What We Know
- Position function: s(t) = 1.5t³ − 13.5t² + 22.5t (ft)
- Initial condition: s = 0 at t = 0 (given by the function itself)
- Time interval: 0 to 6 s
- Find: position at t = 6 s
- Find: total distance traveled over [0, 6 s]
Strategy: Position at t = 6 s is a direct substitution. Total distance is trickier — because s(t) is a cubic, the particle may reverse direction one or more times, making distance ≠ |displacement|. We must differentiate to find v(t), locate any turning points (v = 0) inside [0, 6 s], evaluate s at each turning point, then sum the absolute lengths of each motion segment.
Part A — Position at t = 6 s
Step 1
Substitute t = 6 s Directly into s(t)
The position function is given directly, so no integration is needed — simply substitute t = 6 s:
\[ s(6) = 1.5(6)^3 – 13.5(6)^2 + 22.5(6) \]
\[ = 1.5(216) – 13.5(36) + 135 \]
\[ = 324 – 486 + 135 \]
\[ s(6) = -27 \ \text{ft} \]
Physical meaning: At t = 6 s the particle is 27 ft to the left of the origin. But this tells us nothing about how it got there — it could have gone back and forth multiple times. That is exactly what we investigate in Part B.
Part B — Total Distance Traveled (0 to 6 s)
Step 2
Differentiate s(t) to Find Velocity v(t)
To find where the particle reverses direction, we need the velocity function:
\[ v(t) = \frac{ds}{dt} = 4.5t^2 – 27t + 22.5 \quad \text{ft/s} \]
Why check velocity? A reversal of direction means the velocity crosses zero. If we miss a turning point, we will undercount the total distance — a common and costly error in kinematics problems.
Step 3
Solve v(t) = 0 to Find Turning Points
Setting v(t) = 0 and solving the quadratic:
\[ 4.5t^2 – 27t + 22.5 = 0 \]
Divide through by 4.5 to simplify:
\[ t^2 – 6t + 5 = 0 \quad \Rightarrow \quad (t-1)(t-5) = 0 \]
\[ t = 1 \ \text{s} \quad \text{and} \quad t = 5 \ \text{s} \]
Both roots lie inside [0, 6 s] — so the particle reverses direction twice. This means the interval [0, 6 s] has three distinct motion segments, each of which must be measured separately.
Step 4
Evaluate s(t) at All Critical Instants
We need the particle’s position at t = 0, 1, 5, and 6 s:
\[
s(0) = 1.5(0) – 13.5(0) + 22.5(0) = 0 \ \text{ft}
\]
\[
s(1) = 1.5(1)^3 – 13.5(1)^2 + 22.5(1) = 1.5 – 13.5 + 22.5 = 10.5 \ \text{ft}
\]
\[
s(5) = 1.5(125) – 13.5(25) + 22.5(5) = 187.5 – 337.5 + 112.5 = -37.5 \ \text{ft}
\]
\[
s(6) = -27 \ \text{ft} \quad \text{(from Part A)}
\]
PARTICLE PATH ALONG THE LINE
s = 0 ft
t = 0
t = 0
s = 10.5 ft
t = 1 s ↺
t = 1 s ↺
s = −37.5 ft
t = 5 s ↺
t = 5 s ↺
(leftmost)
s = −27 ft
t = 6 s (end)
t = 6 s (end)
← Negative direction
▸ Origin (s = 0)
Positive direction →
Step 5
Sum Absolute Segment Lengths for Total Distance
Three segments — one for each direction change:
\[
d_\text{total} = \underbrace{|s(1) – s(0)|}_{\text{seg 1}} + \underbrace{|s(5) – s(1)|}_{\text{seg 2}} + \underbrace{|s(6) – s(5)|}_{\text{seg 3}}
\]
\[
= |10.5 – 0| + |-37.5 – 10.5| + |-27 – (-37.5)|
\]
\[
= 10.5 + 48 + 10.5
\]
\[ d_\text{total} = 69 \ \text{ft} \]
Segment breakdown:
Segment 1 (t = 0 to 1 s): particle moves right 10.5 ft (0 → 10.5 ft).
Segment 2 (t = 1 to 5 s): particle reverses, moves left 48 ft (10.5 → −37.5 ft).
Segment 3 (t = 5 to 6 s): particle reverses again, moves right 10.5 ft (−37.5 → −27 ft).
Net displacement = −27 − 0 = −27 ft; total distance = 69 ft. They differ because of the two reversals.
Final Answers
POSITION AT t = 6 s
s = −27 ft (27 ft to the left of the origin)
TOTAL DISTANCE TRAVELED (0 to 6 s)
d = 10.5 + 48 + 10.5 = 69 ft