International Baccalaureate · DP 2: Analysis and Approaches HL

Kinematics

⌚ ~90 min ▢ 9 blocks ✓ 40 exercises ➣ Prerequisites : Applications of Differentiation, Definite Integrals: Areas and Volumes

I Displacement and Velocity

I.1 Displacement

Kinematics is the study of the motion of objects without considering the forces that cause the motion. Calculus describes motion along a straight line formally. We will develop functions for displacement, velocity, and acceleration, and explore the relationships between them.

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Definition — Displacement Function

For an object $P$ in motion along a straight line, the displacement function $s(t)$ gives the position of the object (its signed distance from a fixed origin) at any time $t \ge 0$.

Example

A particle's displacement from an origin O is given by $s(t) = t^2 - 4t + 3$ cm for $0 \le t \le 5$ seconds.

Find the initial displacement.
Find when the particle is at the origin.

Answer

Initial Displacement ($t=0$):
$s(0) = 0^2 - 4(0) + 3 = 3$ cm. The particle starts 3 cm to the right of the origin (in the positive direction).
At the Origin ($s(t)=0$):
$t^2 - 4t + 3 = 0 \implies (t-1)(t-3) = 0$.
The particle is at the origin at $t=1$ s and $t=3$ s.

Skills to practice

Displacement

I.2 Velocity

Definition — Average Velocity

The average velocity of an object moving in the time interval from $t=t_1$ to $t=t_2$ (where $t_1 < t_2$) is given by $$\begin{aligned} \text{average velocity} &= \frac{\text{change in displacement}}{\text{change in time}} \\ &= \frac{s(t_2) - s(t_1)}{t_2 - t_1}. \end{aligned}$$

Definition — Velocity Function

The instantaneous velocity $v(t)$ of an object is the rate of change of its displacement. It is found by differentiating the displacement function $s(t)$: $$ \textcolor{colordef}{v(t) = s'(t) = \frac{\mathrm{d}s}{\mathrm{d}t}} $$

Note

If $v(t) > 0$, the object is moving to the right (in the positive direction).
If $v(t) < 0$, the object is moving to the left (in the negative direction).
If $v(t) = 0$, the object is instantaneously at rest.

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Proposition — Displacement from Velocity

The change in displacement over the interval $[t_1, t_2]$ is the definite integral of the velocity function: $$ s(t_2) - s(t_1) = \int_{t_1}^{t_2} v(t) \,\mathrm{d}t. $$

Proof

The velocity is the derivative of the displacement, $v = s'$, so $s$ is an antiderivative of $v$ on $[t_1, t_2]$. By the Fundamental Theorem of Calculus, $$\begin{aligned}[t] \int_{t_1}^{t_2} v(t)\,\mathrm{d}t &= \big[s(t)\big]_{t_1}^{t_2} && \text{(Fundamental Theorem of Calculus, $s' = v$)}\\ &= s(t_2) - s(t_1). \end{aligned}$$

Skills to practice

Finding Displacement and Velocity
Finding Displacement from Velocity

II Speed and Acceleration

II.1 Speed and Distance

Definition — Speed

The speed of an object is the magnitude of its velocity: $S(t) = |v(t)|$.

Proposition — Distance

The total distance travelled over $[t_1, t_2]$ is the definite integral of the speed: $$ \text{Distance} = \int_{t_1}^{t_2} |v(t)| \,\mathrm{d}t. $$

Proof

We prove the formula when $v$ changes sign at finitely many instants of $[t_1, t_2]$, the case of every motion studied here. Cut the interval at these instants: $t_1 = c_0 < c_1 < \dots < c_k = t_2$, with $v$ of one sign on each piece $[c_{i-1}, c_i]$.

On a piece where $v \ge 0$, the object moves in the positive direction only, so the distance it travels is the change of position: $$\begin{aligned}[t] s(c_i) - s(c_{i-1}) &= \int_{c_{i-1}}^{c_i} v(t)\,\mathrm{d}t && \text{(displacement from velocity)}\\ &= \int_{c_{i-1}}^{c_i} |v(t)|\,\mathrm{d}t && \text{($v = |v|$ on this piece).} \end{aligned}$$
On a piece where $v \le 0$, the object moves in the negative direction only, and the distance is $s(c_{i-1}) - s(c_i) = \displaystyle\int_{c_{i-1}}^{c_i} (-v(t))\,\mathrm{d}t = \int_{c_{i-1}}^{c_i} |v(t)|\,\mathrm{d}t$.

The total distance is the sum of the distances travelled on the pieces: $$\begin{aligned}[t] \text{Distance} &= \sum_{i=1}^{k} \int_{c_{i-1}}^{c_i} |v(t)|\,\mathrm{d}t\\ &= \int_{t_1}^{t_2} |v(t)|\,\mathrm{d}t && \text{(Chasles's relation).} \end{aligned}$$

Method — Finding the Total Distance Travelled with Technology

For an object moving along a straight line with velocity $v(t)$, the displacement from $t = t_1$ to $t = t_2$ is $\displaystyle\int_{t_1}^{t_2} v(t)\,\mathrm{d}t$ and the total distance travelled is $\displaystyle\int_{t_1}^{t_2} |v(t)|\,\mathrm{d}t$. A graphing calculator evaluates both.

Write the integral first: $\displaystyle\int_{t_1}^{t_2} |v(t)|\,\mathrm{d}t$ for the distance, $\displaystyle\int_{t_1}^{t_2} v(t)\,\mathrm{d}t$ for the displacement.
Enter it with the variable $X$ of the calculator in place of $t$, and the absolute-value function for the distance: $\displaystyle\int_{t_1}^{t_2} |v(X)|\,\mathrm{d}X$. The angle unit is set to radians when $v$ contains a trigonometric function.
Or split at the changes of direction: graph $v$, find its zeros between $t_1$ and $t_2$ with the zero tool (the object changes direction where $v$ changes sign), split $[t_1, t_2]$ at these zeros, integrate $v$ over each piece, from $t_1$ to the first zero and from the last zero to $t_2$ included, and add the absolute values of the results.
Check and report: the distance is at least the absolute value of the displacement, and equals it only when the object does not change direction. Report both to 3 significant figures, with their units.

The mock-up shows the velocity $v(t) = 5\sin t - t$ on $0 \le t \le 6$, in the window $-0.5 \le x \le 6.5$, $-11 \le y \le 4$, where it changes sign once, then the distance evaluated in one integral.

Example

A particle moves along a straight line with velocity $v(t) = 5\sin t - t$ metres per second, for $0 \le t \le 6$ seconds.

Find the displacement of the particle between $t = 0$ and $t = 6$.
Find the total distance travelled by the particle between $t = 0$ and $t = 6$.

Answer

The calculator is set to radians, since $v$ contains $\sin$.

The displacement is $\displaystyle\int_0^6 (5\sin t - t)\,\mathrm{d}t$. The template, with the integrand $5\sin X - X$ and the bounds $0$ and $6$, displays $-17.800851$. The displacement is about $-17.8$ m: at $t = 6$, the particle is $17.8$ m from its starting point, on the negative side.
Check with an antiderivative: $\displaystyle\int_0^6 (5\sin t - t)\,\mathrm{d}t = \left[-5\cos t - \frac{t^2}{2}\right]_0^6 = -5\cos 6 - 18 + 5 \approx -17.8$.
The distance is $\displaystyle\int_0^6 |5\sin t - t|\,\mathrm{d}t$. The template, with the integrand $|5\sin X - X|$, displays $29.609835$. The particle travels about $29.6$ m.
Check by splitting: the zero tool displays the zero $X = 2.5957391$ of $v$; $v$ is positive before it and negative after it, so the particle moves forward, then turns back once. The template displays $\displaystyle\int_0^{2.5957391} v(t)\,\mathrm{d}t = 5.9044917$ and $\displaystyle\int_{2.5957391}^{6} v(t)\,\mathrm{d}t = -23.705343$, and $$\begin{aligned}[t] \text{distance} &= |5.9044917| + \left|-23.705343\right| &&\text{(split at the change of direction)}\\ &= 29.609835 \approx 29.6 \text{ m}. \end{aligned}$$ The distance, $29.6$ m, is greater than $\left|-17.8\right| = 17.8$ m, since the particle changes direction.

Skills to practice

Finding the Total Distance
Finding the Total Distance with Technology

II.2 Acceleration

Definition — Acceleration Function

The acceleration $a(t)$ of an object is the rate of change of its velocity. It is the second derivative of the displacement function: $$ \textcolor{colordef}{a(t) = v'(t) = s''(t)} $$

The kinematic functions are related through differentiation and integration (integration recovers $s$ and $v$ only up to a constant, which is fixed by an initial value).

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Proposition — Sign Test for Speed

The speed $|v(t)|$ is increasing (the motion speeds up) when $v(t)$ and $a(t)$ have the same sign.
The speed $|v(t)|$ is decreasing (the motion slows down) when $v(t)$ and $a(t)$ have opposite signs.

Here $v(t)$ is the signed velocity: the test is about its value $|v(t)|$, not about $v(t)$ itself.

Example

The displacement of a particle is given by $s(t) = t^3 - 6t^2 + 9t$ cm, for $t \ge 0$.

Find expressions for the velocity and acceleration.
Find when the particle's speed $|v(t)|$ is increasing.

Answer

Velocity and Acceleration: $$ v(t) = s'(t) = 3t^2 - 12t + 9 \;\text{cm/s}, $$ $$ a(t) = v'(t) = 6t - 12 \;\text{cm/s}^2. $$
Increasing Speed:
We need $v(t)$ and $a(t)$ to have the same sign. First, find the roots: $$ v(t) = 3(t^2-4t+3) = 3(t-1)(t-3) = 0 \implies t=1,\; 3, $$ $$ a(t) = 6(t-2) = 0 \implies t=2. $$ We build a sign table for both functions: Speed is increasing when $v(t)$ and $a(t)$ have the same sign, that is for $$ 1 < t < 2 \quad \text{and} \quad t > 3. $$

Skills to practice

Finding Velocity, Acceleration and Distance
Finding Velocity and Distance from Acceleration

III Going further

Skills to practice

Solving Problems
Writing Mathematics
Quiz