Calculus was invented to describe motion. Newton created it to explain why planets orbit the sun; today, engineers use it to design rockets, autonomous vehicles, and roller coasters. The connections between position, velocity, and acceleration — the fundamental triad of kinematics — are the most direct application of integration and differentiation. In this lesson, we bring together everything from the module: find velocity by integrating acceleration, find displacement by integrating velocity, and use initial conditions to determine constants. This is where calculus stops being abstract and starts describing the physical world.
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A car accelerates from rest with $a(t) = 2t$ m/s². How would you find its velocity after 3 seconds? And how far has it travelled in those 3 seconds?
Core Content
For a particle moving along a straight line:
Going up (integration):
v(t) = ∫ a(t) dt + C_1
x(t) = ∫ v(t) dt + C_2
Going down (differentiation):
v(t) = dx/dt
a(t) = dv/dt = d²x/dt²
Displacement vs Distance:
These are equal only if $v(t) \geq 0$ throughout the interval.
Example 1: A particle has acceleration $a(t) = 6t - 4$ m/s². Initially at rest at the origin, find its position at $t = 2$.
v(t) = ∫(6t - 4)dt = 3t² - 4t + C₁
v(0) = 0: C₁ = 0, so v(t) = 3t² - 4t
x(t) = ∫(3t² - 4t)dt = t³ - 2t² + C₂
x(0) = 0: C₂ = 0, so x(t) = t³ - 2t²
x(2) = 8 - 8 = 0 metres
Example 2: A particle has velocity $v(t) = t^2 - 4t + 3$ m/s. Find the total distance travelled from $t = 0$ to $t = 3$.
v(t) = (t-1)(t-3), so v = 0 at t = 1 and t = 3
On [0,1]: v ≥ 0 (test t = 0.5: 0.25 - 2 + 3 > 0)
On [1,3]: v ≤ 0 (test t = 2: 4 - 8 + 3 < 0)
Distance = ∫₀¹ v dt + ∫₁³ (-v) dt
= [t³/3 - 2t² + 3t]₀¹ + [-t³/3 + 2t² - 3t]₁³
= (1/3 - 2 + 3) + [(-9 + 18 - 9) - (-1/3 + 2 - 3)]
= 4/3 + [0 - (-4/3)] = 4/3 + 4/3 = 8/3 metres
A particle moves with $a(t) = 2$ m/s². At $t = 0$, $v = 5$ m/s and $x = 10$ m. Find $x(t)$.
The position function.
v(t) = ∫2 dt = 2t + C₁
v(0) = 5: C₁ = 5, so v(t) = 2t + 5
x(t) = ∫(2t + 5) dt = t² + 5t + C₂
x(0) = 10: C₂ = 10
x(t) = t² + 5t + 10
$x(t) = t^2 + 5t + 10$ metres.
A particle has $v(t) = 3t^2 - 12t + 9$ m/s. Find the total distance travelled from $t = 0$ to $t = 4$.
Answer:
$v(t) = 3(t-1)(t-3)$. Zeroes at $t = 1, 3$.
On $[0,1]$: $v \geq 0$. On $[1,3]$: $v \leq 0$. On $[3,4]$: $v \geq 0$.
Distance = $\int_0^1 v\,dt + \int_1^3 (-v)\,dt + \int_3^4 v\,dt = 4 + 4 + 4 = 12$ m.
$v = \frac{dx}{dt}$, $a = \frac{dv}{dt}$
$v = \int a\,dt$, $x = \int v\,dt$
$\Delta x = \int_{t_1}^{t_2} v\,dt$
$\int_{t_1}^{t_2} |v(t)|\,dt$
For $a(t) = 2t$ with $v(0) = 0$: $v(t) = \int 2t\,dt = t^2 + C$. Since $v(0) = 0$, $C = 0$, so $v(t) = t^2$. After 3 seconds: $v(3) = 9$ m/s. For displacement: $x(t) = \int t^2\,dt = \frac{t^3}{3} + C_2$. With $x(0) = 0$, $C_2 = 0$, so $x(3) = 9$ metres. Notice how each integration step requires an initial condition to fix the constant. Without $v(0) = 0$, we couldn't determine $C$; without $x(0) = 0$, we couldn't determine $C_2$. This is why second-order motion problems need two initial conditions.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. A particle has acceleration $a(t) = 6 - 2t$ m/s². At $t = 0$, $v = 1$ m/s and $x = 2$ m. Find $v(t)$ and $x(t)$, then evaluate $x(3)$. Show all working. 3 MARKS
9. A particle moves with velocity $v(t) = t^2 - 5t + 6$ m/s. Find: (a) when the particle is at rest, (b) the displacement from $t = 0$ to $t = 4$, (c) the total distance travelled from $t = 0$ to $t = 4$. Show all working. 4 MARKS
10. In autonomous vehicle navigation, accelerometers measure $a(t)$ and the computer must determine position $x(t)$. (a) Explain why two integrations are needed. (b) What errors can accumulate if the accelerometer has small measurement errors? (c) How do GPS systems correct these errors? 3 MARKS
1. $v(t) = 2t^2 - 2t + 3$, $x(t) = \frac{2t^3}{3} - t^2 + 3t$. $x(2) = \frac{16}{3} - 4 + 6 = \frac{22}{3}$ m.
2. Stops when $v = 0$: $12 - 2t = 0$, $t = 6$ s. Distance = $\int_0^6 (12-2t)\,dt = [12t - t^2]_0^6 = 72 - 36 = 36$ m.
3. $v(t) = -10t + 20$. Max height when $v = 0$: $t = 2$ s. $x(t) = -5t^2 + 20t$. $x(2) = -20 + 40 = 20$ m.
1. Displacement is net change (can be negative). Distance is total path length (always positive). Example: walking 3m forward then 2m back gives displacement 1m, distance 5m.
2. The particle may have moved away and returned. Distance measures the full path; displacement only measures the endpoints.
3. GPS integrates accelerometer data (twice) to estimate position, then corrects using satellite position fixes to eliminate drift from integration errors.
Q8 (3 marks): $v(t) = 6t - t^2 + C_1$ [0.5]. $v(0) = 1$: $C_1 = 1$, so $v(t) = 6t - t^2 + 1$ [0.5]. $x(t) = 3t^2 - \frac{t^3}{3} + t + C_2$ [0.5]. $x(0) = 2$: $C_2 = 2$ [0.5]. $x(3) = 27 - 9 + 3 + 2 = 23$ m [1].
Q9 (4 marks): (a) $v(t) = (t-2)(t-3) = 0$, so $t = 2, 3$ s [1]. (b) Displacement = $\int_0^4 (t^2 - 5t + 6)\,dt = [\frac{t^3}{3} - \frac{5t^2}{2} + 6t]_0^4 = \frac{64}{3} - 40 + 24 = \frac{64}{3} - 16 = \frac{16}{3}$ m [1.5]. (c) On $[0,2]$: $v \geq 0$; on $[2,3]$: $v \leq 0$; on $[3,4]$: $v \geq 0$. Distance = $\int_0^2 v\,dt + \int_2^3 (-v)\,dt + \int_3^4 v\,dt = \frac{14}{3} + \frac{1}{6} + \frac{5}{6} = \frac{14}{3} + 1 = \frac{17}{3}$ m [1.5].
Q10 (3 marks): (a) Acceleration → velocity → position requires two integrations [1]. (b) Small acceleration errors integrate to velocity errors, which integrate again to position errors — errors accumulate quadratically over time (drift) [1]. (c) GPS provides absolute position fixes that reset the accumulated error, preventing the drift that pure inertial navigation would suffer [1].
Climb platforms by integrating acceleration to find velocity and position, and calculating total distance travelled. Pool: lesson 14.
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