Kinematics

Displacement, Velocity, and Acceleration (CED Unit 1)

Kinematics describes the motion of objects without considering the forces that cause motion.

Scalar vs Vector Quantities

Quantity	Type	Description
Displacement	Vector	Change in position, $\Delta \vec{r}$
Distance	Scalar	Total path length traveled
Velocity	Vector	Rate of change of displacement
Speed	Scalar	Rate of change of distance (magnitude of velocity)
Acceleration	Vector	Rate of change of velocity

Definitions Using Calculus

For position $\vec{r}(t)$:

$$\vec{v}(t) = \frac{d\vec{r}}{dt}$$ $$\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}$$

Conversely, by integration:

$$\vec{v}(t) = \vec{v}_0 + \int_0^t \vec{a}(t')\, dt'$$ $$\vec{r}(t) = \vec{r}_0 + \int_0^t \vec{v}(t')\, dt'$$

Why Calculus Is the Natural Language of Kinematics

Velocity is the derivative of position. Acceleration is the derivative of velocity. These are not Empirical observations; they are the definitions of velocity and acceleration. The kinematic Equations are the results of integrating constant acceleration once (to get velocity) and Twice (to get position). Every result in kinematics follows from these definitions plus the Assumption of constant acceleration. If you understand the calculus, you understand the kinematics.

One-Dimensional Kinematics

Explore the simulation above to develop intuition for this topic.

Constant Acceleration Equations

When acceleration $a$ is constant, integrating twice yields:

$$V(t) = v_0 + at$$ $$X(t) = x_0 + v_0 t + \frac{1}{2}at^2$$ $$V^2 = v_0^2 + 2a(x - x_0)$$ $$X - x_0 = \frac{v_0 + v}{2} t$$

Derivation from Calculus

Starting from a = \frac{dv}{dt} = \mathrm{const:

$$\int_{v_0}^{v} dv' = \int_0^t a\, dt' \implies v - v_0 = at \implies v = v_0 + at$$

Then $v = \frac{dx}{dt}$:

$$\int_{x_0}^{x} dx' = \int_0^t (v_0 + at')\, dt' \implies x - x_0 = v_0 t + \frac{1}{2}at^2$$

Eliminating Time: The $v^2$ Equation

To eliminate $t$ from the first two equations, substitute $t = (v - v_0)/a$ into the position Equation:

$$X - x_0 = v_0 \frac{v - v_0}{a} + \frac{1}{2}a\left(\frac{v - v_0}{a}\right)^2$$ $$X - x_0 = \frac{v_0 v - v_0^2}{a} + \frac{v^2 - 2v_0 v + v_0^2}{2a}$$ $$2a(x - x_0) = 2v_0 v - 2v_0^2 + v^2 - 2v_0 v + v_0^2 = v^2 - v_0^2$$ $$V^2 = v_0^2 + 2a(x - x_0)$$

This equation is useful because it relates velocity and displacement without requiring the time.

When to Use Each Equation

Each of the four kinematic equations relates a different subset of the five variables ($v$, $v_0$ $a$, $t$, $x - x_0$). Choose the equation that contains the variables you know and the variable you Need to find:

Known variables	Use equation
$v_0$, $a$, $t$	$v = v_0 + at$
$v_0$, $a$, $t$	$x - x_0 = v_0 t + \frac{1}{2}at^2$
$v_0$, $a$, $\Delta x$	$v^2 = v_0^2 + 2a\Delta x$
$v_0$, $v$, $t$	$\Delta x = \frac{v_0 + v}{2} t$

The variable that does not appear in each equation tells you when to use it.

Free Fall

Near Earth’s surface, the acceleration due to gravity is approximately:

G \approx 9.8 \mathrm{ m/s^2 \approx 32 \mathrm{ ft/s^2

Directed downward. In the standard coordinate system (upward positive), $a = -g$.

Equations of Free Fall

$$V = v_0 - gt$$ $$Y = y_0 + v_0 t - \frac{1}{2}gt^2$$ $$V^2 = v_0^2 - 2g(y - y_0)$$

Symmetry of Free Fall

An object thrown upward and an object dropped from the same height take the same time to reach the Ground (if air resistance is negligible). This is because the motion is symmetric about the peak: The time to rise equals the time to fall, and the speed at any height on the way up equals the speed At the same height on the way down. The symmetry is a direct consequence of the equations being Quadratic in $t$.

Two-Dimensional Kinematics (Projectile Motion) (CED Unit 1)

In projectile motion, the horizontal and vertical motions are independent.

Why Horizontal and Vertical Motions Are Independent

The horizontal component of velocity has no effect on the vertical acceleration (which is always $g$Downward), and vice versa. This is because the gravitational force acts only vertically. There Is no horizontal force (ignoring air resistance), so the horizontal acceleration is zero. The two Components of motion are completely decoupled, and can be analyzed separately using the same Kinematic equations.

Horizontal Motion (constant velocity, no acceleration)

$$X(t) = x_0 + v_{0x} t = x_0 + (v_0 \cos\theta) t$$ V_x(t) = v_0 \cos\theta \quad \mathrm{(constant)

Vertical Motion (free fall)

$$Y(t) = y_0 + v_{0y} t - \frac{1}{2}gt^2 = y_0 + (v_0 \sin\theta) t - \frac{1}{2}gt^2$$ $$V_y(t) = v_0 \sin\theta - gt$$

Key Results

• Time to reach maximum height: \displaystyle t_{\mathrm{up} = \frac{v_0 \sin\theta}{g}
• Maximum height: $\displaystyle H = \frac{(v_0 \sin\theta)^2}{2g}$
• Total time of flight: $\displaystyle T = \frac{2v_0 \sin\theta}{g}$
• Range: $\displaystyle R = \frac{v_0^2 \sin 2\theta}{g}$
• Maximum range (for fixed $v_0$): achieved at $\theta = 45^\circ$

Proof of the Range Formula

The projectile lands when $y = 0$:

$$0 = (v_0 \sin\theta) T - \frac{1}{2}gT^2 \implies T = \frac{2v_0 \sin\theta}{g}$$

The horizontal distance at time $T$:

$$R = (v_0 \cos\theta) T = (v_0 \cos\theta) \cdot \frac{2v_0 \sin\theta}{g} = \frac{v_0^2 \cdot 2\sin\theta \cos\theta}{g} = \frac{v_0^2 \sin 2\theta}{g}$$

Using the double-angle identity $\sin 2\theta = 2\sin\theta \cos\theta$.

Why Maximum Range Is at 45 Degrees

The range $R = (v_0^2/g)\sin 2\theta$ is maximized when $\sin 2\theta$ is maximized, i.e., when $2\theta = 90^{\circ}$Giving $\theta = 45^{\circ}$. Complementary angles give the same range: $\sin 2\theta = \sin 2(90^{\circ} - \theta)$. A projectile launched at $30^{\circ}$ and one launched At $60^{\circ}$ with the same speed will travel the same horizontal distance, but the $60^{\circ}$ Launch goes higher and stays in the air longer.

The Trajectory Is a Parabola

Eliminating $t$ from the horizontal and vertical equations:

$$T = \frac{x}{v_0 \cos\theta}$$ $$Y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}$$

This is a quadratic in $x$So the trajectory is a parabola. The negative coefficient of $x^2$ Confirms that the parabola opens downward.

Projectile from a Height

When a projectile is launched from a height $h$ above the landing level, the time of flight is Longer than $2v_0\sin\theta/g$ because the projectile must travel further vertically. There is no Simple closed-form formula for the range in this case; instead, find $t$ from the quadratic $y(t) = 0$Then compute $R = v_0\cos\theta \cdot t$.

Position, Velocity, and Acceleration as Graphs (CED Unit 1)

From Graphs

• The slope of a position-time graph gives velocity.
• The slope of a velocity-time graph gives acceleration.
• The area under a velocity-time graph gives displacement.
• The area under an acceleration-time graph gives the change in velocity.

Qualitative Analysis

• Positive slope on $x$-vs-$t$: moving in the positive direction.
• Zero slope on $x$-vs-$t$: at rest.
• Negative slope on $x$-vs-$t$: moving in the negative direction.
• Curvature on $x$-vs-$t$: indicates acceleration (concave up = positive $a$Concave down = negative $a$).

Inflection Points and Velocity Extrema

On an $x$-vs-$t$ graph, an inflection point (where the curvature changes sign) corresponds to zero Acceleration. This is the point where the velocity graph crosses its extremum (maximum or minimum). At this instant, the object is changing from speeding up to slowing down (or vice versa).

Uniform Circular Motion (CED Unit 3)

An object moving in a circle of radius $r$ at constant speed $v$ has:

• Centripetal acceleration: $\displaystyle a_c = \frac{v^2}{r}$ directed toward the center of the circle.
• Period: $\displaystyle T = \frac{2\pi r}{v}$
• Frequency: $\displaystyle f = \frac{1}{T} = \frac{v}{2\pi r}$

Angular Quantities

$$\omega = \frac{d\theta}{dt} = \frac{v}{r}$$ $$A_c = \omega^2 r = \frac{v^2}{r}$$

Derivation of Centripetal Acceleration

Consider a particle moving at constant speed $v$ on a circle of radius $r$. Over a small time Interval $\Delta t$The velocity vector changes direction by angle $\Delta\theta$ while maintaining Magnitude $v$.

The change in velocity $\Delta\vec{v}$ is perpendicular to $\vec{v}$ and has magnitude:

$$|\Delta\vec{v}| = v \cdot \Delta\theta$$

Since $\Delta\theta = \frac{v \Delta t}{r}$:

$$|\Delta\vec{v}| = v \cdot \frac{v \Delta t}{r} = \frac{v^2 \Delta t}{r}$$

The centripetal acceleration is:

$$A_c = \lim_{\Delta t \to 0} \frac{|\Delta\vec{v}|}{\Delta t} = \frac{v^2}{r}$$

Why Centripetal Acceleration Does Not Change Speed

Centripetal acceleration is always perpendicular to the velocity. Since the dot product $\vec{a}_c \cdot \vec{v} = 0$The power delivered is $P = \vec{F} \cdot \vec{v} = 0$And the Kinetic energy $\frac{1}{2}mv^2$ is constant. The speed does not change; only the direction of the Velocity vector changes. This is why “centripetal” means “centre-seeking” — the acceleration always Points toward the center, never along the velocity.

Non-Uniform Acceleration (AP Physics C)

When acceleration is not constant, use calculus.

Velocity from Variable Acceleration

$$V(t) = v_0 + \int_0^t a(t')\, dt'$$

Position from Variable Velocity

$$X(t) = x_0 + \int_0^t v(t')\, dt'$$

Displacement as Area Under the Curve

When $a(t)$ or $v(t)$ is given graphically rather than analytically, the displacement can be found By computing the area under the $v$-vs-$t$ curve. For non-linear curves, this requires numerical Integration (counting squares, the trapezoidal rule, or Simpson’s rule).

Worked Example: Reversing the Problem

A particle has acceleration $a(t) = 12t$ m/s$^2$Initial velocity $v(0) = 5$ m/s, and initial Position $x(0) = 0$. Find $v(t)$ and $x(t)$.

$$V(t) = v(0) + \int_0^t 12t'\, dt' = 5 + 6t^2$$ $$X(t) = x(0) + \int_0^t (5 + 6t'^2)\, dt' = 5t + 2t^3$$

Relative Motion

In one dimension with two reference frames $S$ and $S'$Where $S'$ moves with velocity $v_{S'/S}$ Relative to $S$:

$$\vec{v}_{P/S} = \vec{v}_{P/S'} + \vec{v}_{S'/S}$$

This is the Galilean velocity addition formula.

Relative Motion in Two Dimensions

In two dimensions, the velocity addition is done component by component:

$$V_{P/x} = v_{P/S'}\cos\alpha + v_{S'/S}\cos\beta$$ $$V_{P/y} = v_{P/S'}\sin\alpha + v_{S'/S}\sin\beta$$

Where $\alpha$ is the angle of the particle’s velocity in frame $S'$ and $\beta$ is the angle of Frame $S'$ relative to frame $S$.

Crossing the River Perpendicular to the Bank

To land directly across the river, the boat must angle upstream so that the downstream component of Its velocity relative to the water cancels the river current. If the boat has speed $v_b$ and the River flows at $v_r$:

$$\sin\theta = \frac{v_r}{v_b}$$

Where $\theta$ is the angle upstream from the perpendicular. This requires $v_b \gt v_r$; otherwise, The boat cannot overcome the current.

Common Pitfalls

1. Confusing displacement with distance. Displacement is a vector (net change in position); distance is a scalar (total path length).
2. Ignoring the sign convention. In free fall, choosing “up as positive” means $a = -g$. Be consistent throughout the problem.
3. Treating horizontal and vertical components as dependent. In projectile motion, the horizontal motion has constant velocity (no horizontal acceleration, ignoring air resistance).
4. Using the constant acceleration equations when acceleration is not constant. Use calculus (integration) instead.
5. Confusing speed and velocity. Speed is the magnitude of velocity. An object can have constant speed but changing velocity (e.g., uniform circular motion).
6. Forgetting that centripetal acceleration changes direction, not speed. In uniform circular motion, the speed is constant but the velocity vector changes direction continuously.
7. Incorrectly applying the range formula when the launch and landing heights differ.
8. Forgetting that the kinematic equations assume constant acceleration. If the problem involves variable acceleration (a spring, a pendulum, a rocket losing mass), you must use calculus.
9. Mixing up initial and final values. $v_0$ is the velocity at the start of the interval; $v$ is the velocity at the end. Be careful about which instant each variable refers to.
10. Assuming the maximum height occurs at half the total time. This is only true when the launch and landing heights are the same.

Practice Questions

1. A ball is dropped from a height of 80 m. How long does it take to reach the ground, and what is its speed at impact?

2. A projectile is launched from a cliff 60 m high at 30 \mathrm{ m/s at $40^\circ$ above horizontal. Find the time of flight and the horizontal range.

3. A particle has acceleration a(t) = 6t \mathrm{ m/s^2 with initial velocity v(0) = -4 \mathrm{ m/s and initial position x(0) = 2 \mathrm{ m. Find $v(t)$, $x(t)$And the displacement from $t = 0$ to $t = 3$.

4. A car accelerates from rest at 2.0 \mathrm{ m/s^2 for 200 m. What is its final velocity?

5. An object moves along the $x$-axis with velocity $v(t) = t^3 - 6t^2 + 9t$. Find the total distance traveled from $t = 0$ to $t = 5$.

6. A satellite orbits Earth at an altitude of 300 km in a circular orbit. If the orbital period is 90.5 minutes, find the orbital speed and centripetal acceleration. (Earth’s radius = $6371$ km.)

7. A swimmer can swim at 2.0 \mathrm{ m/s in still water. She needs to cross a river 50 m wide flowing at 1.5 \mathrm{ m/s. At what angle upstream should she head to land directly across?

8. Derive the formula for the maximum height of a projectile using calculus.

9. A ball is thrown from the top of a 50 m building at 20 \mathrm{ m/s at $30^\circ$ below the horizontal. Find the time to hit the ground and the horizontal distance from the building.

10. The position of a particle is $x(t) = t^3 - 6t^2 + 9t + 1$ m. Find (a) the times when the particle is at rest, (b) the acceleration at each of those times, and (c) the total distance traveled between $t = 0$ and $t = 5$ s.

Motion Graphs: Advanced Analysis

Determining Velocity from a Displacement-Time Graph

For a curved $x$-vs-$t$ graph, the instantaneous velocity at any point is the gradient of the Tangent to the curve at that point. A straight-line section indicates constant velocity (zero Acceleration). A section that curves upward (concave up) indicates positive acceleration; a section That curves downward (concave down) indicates negative acceleration.

Worked Example: Interpreting a Velocity-Time Graph

A particle moves along the $x$-axis. Its velocity is given by $v(t) = 4t - t^2$ for $0 \le t \le 5$ S.

Velocity at $t = 2$ s: $v(2) = 8 - 4 = 4$ m/s.

Acceleration: $a(t) = \frac{dv}{dt} = 4 - 2t$. At $t = 2$: $a = 0$ (turning point of velocity).

Displacement from $t = 0$ to $t = 4$:

\Delta x = \int_0^4 (4t - t^2)\, dt = \left[2t^2 - \frac{t^3}{3}\right]_0^4 = 32 - \frac{64}{3} = 32 - 21.33 = 10.67 \mathrm{ m

Total distance from $t = 0$ to $t = 5$: The velocity is zero at $t = 0$ and $t = 4$. For 0 \lt t \lt 4$$v \gt 0. For 4 \lt t \lt 5$$v \lt 0.

$d = \int_0^4 (4t - t^2)\, dt + \left|\int_4^5 (4t - t^2)\, dt\right| = 10.67 + \left|\left[2t^2 - \frac{t^3}{3}\right]_4^5\right|$

= 10.67 + \left|(50 - 41.67) - (32 - 21.33)\right| = 10.67 + |8.33 - 10.67| = 10.67 + 2.33 = 13.0 \mathrm{ m

Uniform Circular Motion: Extended Analysis

Period, Frequency, and Angular Velocity Relationships

For an object in uniform circular motion:

$\omega = 2\pi f = \frac{2\pi}{T}, \qquad v = \frac{2\pi r}{T} = 2\pi rf, \qquad a_c = \frac{4\pi^2 r}{T^2} = \omega^2 r = \frac{v^2}{r}$

Worked Example: Geostationary Orbit

A geostationary satellite orbits above the equator with a period of 24 hours. Given Earth’s radius $R_E = 6.371 \times 10^6$ m and mass $M_E = 5.97 \times 10^{24}$ kg, find the altitude of the orbit.

T = 86400 \mathrm{ s

$T^2 = \frac{4\pi^2 r^3}{GM}$

$r^3 = \frac{GMT^2}{4\pi^2} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 86400^2}{4\pi^2}$

$r^3 = \frac{2.987 \times 10^{24}}{39.48} = 7.566 \times 10^{22}$

r = \sqrt[3]{7.566 \times 10^{22}} = 4.225 \times 10^7 \mathrm{ m

h = r - R_E = 4.225 \times 10^7 - 6.371 \times 10^6 = 3.588 \times 10^7 \mathrm{ m \approx 35880 \mathrm{ km

This is approximately 36000 km above Earth’s surface, the standard geostationary orbit altitude.

Non-Uniform Acceleration: Advanced Examples (AP Physics C)

Worked Example: Position, Velocity, and Acceleration from a Single Function

The position of a particle is $x(t) = t^4 - 4t^3 + 6t^2$ m.

$v(t) = \frac{dx}{dt} = 4t^3 - 12t^2 + 12t$

$a(t) = \frac{dv}{dt} = 12t^2 - 24t + 12 = 12(t^2 - 2t + 1) = 12(t-1)^2$

The acceleration is always non-negative (it is zero at $t = 1$). The particle is never decelerating.

Times when at rest: $v(t) = 0 \implies 4t(t^2 - 3t + 3) = 0$. The quadratic $t^2 - 3t + 3 = 0$ Has discriminant $9 - 12 = -3 \lt 0$So the only solution is $t = 0$.

The particle is at rest only at $t = 0$.

Worked Example: Piecewise-Defined Acceleration

A particle starts from rest. For $0 \le t \le 3$ s, $a = 6t$ m/s$^2$. For $t \gt 3$ s, $a = 0$.

Phase 1 ($0 \le t \le 3$):

$v(t) = \int_0^t 6t'\, dt' = 3t^2$

$x(t) = \int_0^t 3t'^2\, dt' = t^3$

At $t = 3$: $v = 27$ m/s, $x = 27$ m.

Phase 2 ($t \gt 3$): Constant velocity $v = 27$ m/s.

$x(t) = 27 + 27(t - 3)$

Summary Table: Kinematic Equations

Equation	Missing Variable	Best Used When
$v = v_0 + at$	$x - x_0$	You know initial velocity, acceleration, time
$x - x_0 = v_0 t + \frac{1}{2}at^2$	$v$	You know initial velocity, acceleration, time
$v^2 = v_0^2 + 2a(x - x_0)$	$t$	You know velocities and displacement
$x - x_0 = \frac{v_0 + v}{2}t$	$a$	You know velocities and time (uniform acceleration)

Common Pitfalls: Extended

11. Assuming that the magnitude of acceleration equals $g$ for objects thrown upward. The magnitude is $g$ only when air resistance is negligible. With air resistance, the acceleration while going up is greater than $g$ (drag and gravity both act downward) and while coming down is less than $g$ (drag opposes gravity).

12. Using the wrong reference frame in relative motion problems. Always identify the observer’s frame and the moving frame explicitly. Write the velocity addition formula carefully.

13. Forgetting that in 2D projectile motion, the time determines everything. Once you find the time of flight (from the vertical motion), you can find the horizontal range. You cannot find the range without first finding the time.

14. Confusing angular displacement with linear displacement in circular motion problems. Angular displacement $\theta$ is dimensionless (radians). Linear displacement $s = r\theta$ has units of metres.

15. Assuming that $a = v^2/r$ applies when the speed is not constant. The formula $a_c = v^2/r$ gives only the centripetal component of acceleration. If the speed is changing, there is also a tangential component $a_t = dv/dt$.

Practice Questions (Additional)

11. A particle moves along the $x$-axis with acceleration $a(t) = 8t - 6$ m/s$^2$. At $t = 0$The particle is at $x = 2$ m with velocity $v = -3$ m/s. Find (a) the time when the particle is at rest, (b) the position at that time, and (c) the total distance traveled between $t = 0$ and $t = 3$ s.

12. A projectile is launched from ground level at 40 \mathrm{ m/s. Find the two launch angles that give a range of 120 \mathrm{ m.

13. A car accelerates from rest at 3 \mathrm{ m/s^2 for 4 \mathrm{ sThen travels at constant speed for 6 \mathrm{ sThen decelerates uniformly to rest in 3 \mathrm{ s. Find the total distance and draw a velocity-time graph.

14. Two boats leave the same point simultaneously. Boat A heads north at 6 \mathrm{ m/s and Boat B heads northeast at 8 \mathrm{ m/s. Find the velocity of Boat A relative to Boat B.

15. An astronaut on the Moon throws a ball vertically upward at 15 \mathrm{ m/s. Find the maximum height and total time in the air. (g_{\mathrm{Moon} = 1.62 \mathrm{ m/s^2.)

Extended Worked Examples

Example 16: Relative Motion in Two Dimensions

Airplane A flies north at 250 \mathrm{ m/s. Airplane B flies at 200 \mathrm{ m/s on a heading of $60^\circ$ east of north. Find the velocity of A relative to B.

Step 1: Write velocity vectors

\vec{v}_A = 250\hat{j} \mathrm{ m/s

\vec{v}_B = 200\sin 60°\hat{i} + 200\cos 60°\hat{j} = 173.2\hat{i} + 100\hat{j} \mathrm{ m/s

Step 2: Relative velocity

\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = -173.2\hat{i} + (250 - 100)\hat{j} = -173.2\hat{i} + 150\hat{j} \mathrm{ m/s

Step 3: Magnitude and direction

|\vec{v}_{AB}| = \sqrt{173.2^2 + 150^2} = \sqrt{29998 + 22500} = \sqrt{52498} = 229.1 \mathrm{ m/s

\theta = \arctan\left(\frac{150}{-173.2}\right) = 180° - 40.9° = 139.1° \mathrm{ from east

So the velocity of A relative to B is 229.1 \mathrm{ m/s at $139.1^\circ$ from east (or $50.9^\circ$ west of North).

Example 17: Non-Uniform Acceleration from a Velocity Function

The velocity of a particle is given by $v(t) = t^3 - 6t^2 + 9t$ (m/s) for 0 \le t \le 5 \mathrm{ s.

Find (a) when the particle is at rest, (b) the total distance travelled, and (c) the displacement.

Step 1: Find when the particle is at rest

$v(t) = t^3 - 6t^2 + 9t = t(t^2 - 6t + 9) = t(t - 3)^2 = 0$

$t = 0$ or t = 3 \mathrm{ s.

Step 2: Determine the sign of $v$ between critical points

• $0 \lt t \lt 3$: Test $t = 1$: $v = 1 - 6 + 9 = 4 \gt 0$ (moving forward)
• $t \gt 3$: Test $t = 4$: $v = 64 - 96 + 36 = 4 \gt 0$ (moving forward)

The particle never moves backward. The velocity is always non-negative (zero only at $t = 0$ and $t = 3$).

Step 3: Total distance = displacement (since $v \ge 0$ always)

$s(5) - s(0) = \int_0^5 (t^3 - 6t^2 + 9t) \, dt = \left[ \frac{t^4}{4} - 2t^3 + \frac{9t^2}{2} \right]_0^5$

= \frac{625}{4} - 250 + \frac{225}{2} = 156.25 - 250 + 112.5 = 18.75 \mathrm{ m

Step 4: Acceleration at key points

$a(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t - 1)(t - 3)$

At $t = 0$: a = 9 \mathrm{ m/s^2 (speeding up) At $t = 3$: a = 3(9 - 12 + 3) = 0 \mathrm{ m/s^2 (momentarily stationary, inflection point)

:::info Even though the particle is stationary at t = 3 \mathrm{ sThe acceleration is also zero There, so the particle does not reverse direction. Compare this to projectile motion at the top of The trajectory where $v = 0$ but $a = g \ne 0$So the projectile immediately starts moving Downward.

Example 18: Kinematics with Air Resistance

A 70 \mathrm{ kg skydiver falls from rest. The air resistance force is $F_D = kv^2$ where k = 0.25 \mathrm{ kg/m. Calculate (a) the terminal velocity, (b) the velocity after 10 \mathrm{ s And (c) the distance fallen after 10 \mathrm{ s.

Step 1: Terminal velocity

At terminal velocity: $mg = kv_T^2$

v_T = \sqrt{\frac{mg}{k}} = \sqrt{\frac{70 \times 9.8}{0.25}} = \sqrt{2744} = 52.4 \mathrm{ m/s

Step 2: Analytical solution

The equation of motion is $m\frac{dv}{dt} = mg - kv^2$.

This has the solution:

$v(t) = v_T \tanh\left(\frac{gt}{v_T}\right)$

$v(10) = 52.4 \times \tanh\left(\frac{9.8 \times 10}{52.4}\right) = 52.4 \times \tanh(1.870)$

$\tanh(1.870) = \frac{e^{3.74} - e^{-3.74}}{e^{3.74} + e^{-3.74}} = \frac{42.10 - 0.0238}{42.10 + 0.0238} = 0.9989$

v(10) = 52.4 \times 0.9989 = 52.3 \mathrm{ m/s

The skydiver has essentially reached terminal velocity after 10 \mathrm{ s.

Step 3: Distance fallen

$y(t) = \frac{v_T^2}{g}\ln\left(\cosh\left(\frac{gt}{v_T}\right)\right)$

y(10) = \frac{2744}{9.8}\ln(\cosh(1.870)) = 280 \times \ln(3.263) = 280 \times 1.183 = 331.2 \mathrm{ m

Common Pitfalls Extended

Pitfall 6: Using the Wrong Sign Convention in Free Fall

Always define your positive direction before solving. If upward is positive, then $a = -g$ and Displacements above the launch point are positive. Many errors come from mixing sign conventions Mid-problem, especially when combining horizontal and vertical components.

Pitfall 7: Confusing Average Velocity with Average Speed

Average velocity = \frac{\mathrm{displacement}{\mathrm{time} (a vector, can be zero for round Trips). Average speed = \frac{\mathrm{total distance}{\mathrm{time} (a scalar, always positive). They are equal only for motion in one direction without reversing.

Pitfall 8: Assuming Graph Slopes Give Instantaneous Values Everywhere

The slope of a displacement-time graph gives instantaneous velocity. The slope of a velocity-time Graph gives instantaneous acceleration. However, for curved graphs, the slope changes continuously. You must draw a tangent at the specific point to find the instantaneous value — the average slope Between two points is not the same as the instantaneous slope.

Additional Practice Problems

16. A ball is thrown upward from the top of a 50 \mathrm{ m building at 20 \mathrm{ m/s. Calculate (a) the maximum height above ground, (b) the time to reach maximum height, (c) the total time in the air, and (d) the velocity just before impact with the ground.

17. A police car chasing a speeding motorist. The motorist passes at 30 \mathrm{ m/s. The police car starts 2 \mathrm{ s later with acceleration 3 \mathrm{ m/s^2. Calculate (a) when and where the police car catches the motorist, and (b) the speeds of both vehicles at that moment.

18. A projectile is launched from a cliff of height 80 \mathrm{ m at 50 \mathrm{ m/s at $40^\circ$ above the horizontal. Calculate (a) the time of flight, (b) the maximum height above the launch point, (c) the horizontal range, and (d) the velocity (magnitude and direction) at impact.

19. Two trains approach each other on parallel tracks. Train A travels at 60 \mathrm{ m/s and Train B at 40 \mathrm{ m/s. A bird flies at 80 \mathrm{ m/s from Train A to Train B and back repeatedly until the trains meet. If the trains are initially 2 \mathrm{ km apart, calculate the total distance the bird flies.

20. The position of a particle is given by $x(t) = 2t^3 - 9t^2 + 12t + 1$ (m). Find (a) the times when the particle changes direction, (b) the total distance travelled between $t = 0$ and t = 4 \mathrm{ sAnd (c) the average velocity and average speed over this interval.

Practice Problems

Question 1: Two-object kinematics with relative motion

A ball is thrown vertically upward from ground level with initial velocity 20 \mathrm{ m/s. One second later, a second ball is dropped from a height of 50 \mathrm{ m. At what height above the ground do the two balls meet? Take g = 9.8 \mathrm{ m/s^2.

Answer

Ball 1 (thrown up at $t = 0$): $y_1 = 20t - 4.9t^2$.

Ball 2 (dropped at $t = 1$): $y_2 = 50 - 4.9(t - 1)^2$ for $t \ge 1$.

Set $y_1 = y_2$: $20t - 4.9t^2 = 50 - 4.9(t^2 - 2t + 1) = 50 - 4.9t^2 + 9.8t - 4.9 = 45.1 - 4.9t^2 + 9.8t$.

$20t = 45.1 + 9.8t$So $10.2t = 45.1$, t = 4.42 \mathrm{ s.

Height: y_1 = 20(4.42) - 4.9(4.42)^2 = 88.4 - 95.7 = -7.3 \mathrm{ m.

This is negative, meaning the balls don’t meet before ball 1 hits the ground. Ball 1 lands when $y_1 = 0$: $20t - 4.9t^2 = 0$, t = 20/4.9 = 4.08 \mathrm{ s.

Since $4.42 > 4.08$Ball 1 has already landed. The balls do not meet in the air.

Question 2: Projectile motion with angle optimization

A football is kicked from ground level with speed 25 \mathrm{ m/s. At what angle should it be kicked to maximise the range? Calculate the maximum range and the time of flight.

Answer

For maximum range on level ground, the launch angle is $45^\circ$.

Range: R = \frac{v^2 \sin(2\theta)}{g} = \frac{25^2 \sin(90^\circ)}{9.8} = \frac{625}{9.8} = 63.8 \mathrm{ m.

Time of flight: T = \frac{2v \sin\theta}{g} = \frac{2 \times 25 \times \sin(45^\circ)}{9.8} = \frac{35.36}{9.8} = 3.61 \mathrm{ s.

Question 3: Kinematics with calculus

The velocity of a particle is given by v(t) = 6t^2 - 18t + 12 \mathrm{ m/s for 0 \le t \le 4 \mathrm{ s. Find the total distance travelled and the displacement.

Answer

Set $v = 0$: 6t^2 - 18t + 12 = 0$$t^2 - 3t + 2 = 0$$(t-1)(t-2) = 0. Direction changes at $t = 1$ and $t = 2$.

Acceleration: $a = dv/dt = 12t - 18$.

Displacement = \int_0^4 v \, dt = [2t^3 - 9t^2 + 12t]_0^4 = 128 - 144 + 48 = 32 \mathrm{ m.

Total distance = $\int_0^1 v \, dt + |\int_1^2 v \, dt| + \int_2^4 v \, dt$.

$\int_0^1 = [2 - 9 + 12] = 5$. $\int_1^2 = [16 - 36 + 24] - [2 - 9 + 12] = 4 - 5 = -1$Absolute value = 1. $\int_2^4 = [128 - 144 + 48] - [16 - 36 + 24] = 32 - 44 = -12$Absolute value = 12.

Total distance = 5 + 1 + 12 = 18 \mathrm{ m.

Displacement = 32 m, total distance = 18 m.

Question 4: Free fall with air resistance (conceptual)

Explain qualitatively how the motion of a falling object differs with and without air resistance. Sketch velocity-time graphs for both cases and label the terminal velocity.

Answer

Without air resistance: Velocity increases linearly with time ($v = gt$). The graph is a straight line through the origin.

With air resistance: As the object falls, air resistance (drag) increases with speed. Net force = weight - drag decreases. Acceleration decreases. Eventually, drag equals weight (net force = 0), and the object reaches terminal velocity ($v_T$). The velocity-time graph curves upward, approaching $v_T$ asymptotically. It is concave down, starting steep and becoming flat.

Terminal velocity depends on mass, cross-sectional area, drag coefficient, and air density. A heavier object with smaller area reaches a higher terminal velocity.

Question 5: Relative motion in two dimensions

A boat can travel at 5 \mathrm{ m/s in still water. It needs to cross a river that is 100 \mathrm{ m wide, flowing at 3 \mathrm{ m/s. If the boat heads directly across the river, how far downstream is it carried? What heading should the boat take to land directly across from the starting point?

Answer

Heading directly across: Time to cross = 100 / 5 = 20 \mathrm{ s. Downstream distance = 3 \times 20 = 60 \mathrm{ m.

To land directly across: The boat must angle upstream so its upstream component cancels the current. Let $\theta$ be the angle upstream from the line perpendicular to the bank.

$5\sin\theta = 3$So \sin\theta = 3/5 = 0.6$$\theta = 36.9^\circ.

Velocity across river: 5\cos\theta = 5 \times 0.8 = 4 \mathrm{ m/s. Time to cross: 100 / 4 = 25 \mathrm{ s.

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