Rectilinear Kinematics
Displacement, Average Velocity & Acceleration — Quadratic Position
A particle moves along a straight line with its position defined by
s = (10t² + 20) mm, where t is in seconds.
Determine (a) the displacement of the particle during the interval
from t = 1 s to t = 5 s,
(b) the average velocity over this interval, and
(c) the acceleration of the particle at t = 1 s.
What We Know
- Position function: s(t) = 10t² + 20 (mm)
- Time interval of interest: t = 1 s to t = 5 s
- Find (a): displacement Δs over [1 s, 5 s]
- Find (b): average velocity over [1 s, 5 s]
- Find (c): acceleration at t = 1 s
Strategy:
(a) Evaluate s(t) at both endpoints and subtract: Δs = s(5) − s(1).
(b) Average velocity = Δs / Δt — uses displacement, not total distance.
(c) Differentiate s(t) twice to get a(t); evaluate at t = 1 s. Since s(t) is quadratic, the acceleration will be a constant — no need to check for turning points for parts (a) and (b), as v(t) = 20t > 0 for all t > 0.
Part A — Displacement from t = 1 s to t = 5 s
Step 1
Evaluate s(t) at Both Endpoints
Substitute t = 1 s into the position function:
\[ s(1) = 10(1)^2 + 20 = 10 + 20 = 30 \ \text{mm} \]
Substitute t = 5 s into the position function:
\[ s(5) = 10(5)^2 + 20 = 10(25) + 20 = 250 + 20 = 270 \ \text{mm} \]
Displacement is the change in position:
\[ \Delta s = s(5) – s(1) = 270 – 30 \]
\[ \Delta s = 240 \ \text{mm} \]
No turning-point check needed: The velocity v(t) = 20t is always positive for t > 0, meaning the particle moves in one direction only throughout [1, 5 s]. Displacement and distance are therefore identical here — both equal 240 mm.
Part B — Average Velocity from t = 1 s to t = 5 s
Step 2
Compute Average Velocity
Average velocity is defined as the net displacement divided by the elapsed time:
\[ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{s(5) – s(1)}{5 – 1} = \frac{240 \ \text{mm}}{4 \ \text{s}} \]
\[ \bar{v} = 60 \ \text{mm/s} \]
Average vs. instantaneous velocity: The instantaneous velocity at t = 1 s is v(1) = 20(1) = 20 mm/s, and at t = 5 s it is v(5) = 20(5) = 100 mm/s. The average velocity of 60 mm/s lies exactly halfway between these — which makes sense because the acceleration is constant, meaning velocity increases linearly and the average equals the midpoint value.
Part C — Acceleration at t = 1 s
Step 3
Differentiate s(t) Twice to Find Acceleration
First differentiation gives velocity:
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(10t^2 + 20) = 20t \quad \text{mm/s} \]
Second differentiation gives acceleration:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(20t) = 20 \ \text{mm/s}^2 \]
\[ a = 20 \ \text{mm/s}^2 \quad \text{(constant — the same at } t = 1 \text{ s or any other time)} \]
| Quantity | Expression | At t = 1 s | At t = 5 s |
|---|---|---|---|
| Position s | 10t² + 20 mm | 30 mm | 270 mm |
| Velocity v | 20t mm/s | 20 mm/s | 100 mm/s |
| Acceleration a | 20 mm/s² | 20 mm/s² | 20 mm/s² |
Key insight — the kinematic hierarchy: Because s(t) is quadratic (degree 2), differentiating once gives a linear velocity function, and differentiating again yields a constant acceleration. This is the hallmark of uniformly accelerated motion. The constant 20 mm/s² means velocity grows by exactly 20 mm/s for every second that passes — which is why v = 20 mm/s at t = 1 s and v = 100 mm/s at t = 5 s (a difference of 80 mm/s over 4 s, consistent with 4 × 20 = 80). ✓
Final Answers
(A) DISPLACEMENT — t = 1 s to t = 5 s
Δs = 240 mm
(B) AVERAGE VELOCITY — t = 1 s to t = 5 s
v̄ = 60 mm/s
(C) ACCELERATION AT t = 1 s
a = 20 mm/s² (constant throughout)