Kinematics (Extended)

Kinematics (Extended Treatment)

This document extends the core kinematics material with deeper derivations, multi-stage problems, projectile motion in two dimensions, and the calculus approach to variable acceleration.

info

This page complements the core kinematics notes. Readers should already be comfortable with the SUVAT equations and basic calculus definitions of velocity and acceleration.

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1. Equations of Motion (SUVAT) -- Rigorous Derivation

1.1 Derivation from first principles

Starting from the definition of constant acceleration:

$a = \frac{dv}{dt}$

Integrating with respect to $t$:

$\int_0^t a\,dt = \int_u^v dv \implies at = v - u \implies v = u + at$

This is SUVAT equation 1. We now derive the remaining four.

Equation 2: $s = ut + \tfrac{1}{2}at^2$

Since $v = \dfrac{ds}{dt} = u + at$, integrate:

$\int_0^s ds = \int_0^t (u + at)\,dt \implies s = ut + \frac{1}{2}at^2$

Equation 3: $s = \dfrac{1}{2}(u + v)t$

Substitute $v = u + at$ into Equation 2:

$s = ut + \frac{1}{2}(v - u)t = \frac{1}{2}(2u + v - u)t = \frac{1}{2}(u + v)t$

Equation 4: $v^2 = u^2 + 2as$

From $v = u + at$, square both sides:

$v^2 = (u + at)^2 = u^2 + 2uat + a^2t^2$

Factor $2a$ from the last two terms:

$v^2 = u^2 + 2a\!\left(ut + \frac{1}{2}at^2\right) = u^2 + 2as$

Equation 5: $s = vt - \tfrac{1}{2}at^2$

Substitute $u = v - at$ into Equation 2:

$s = (v - at)t + \frac{1}{2}at^2 = vt - \frac{1}{2}at^2$

1.2 Selecting the correct equation

The key skill is identifying which variable is unknown and which is not needed:

Unknown	Do not use
displacement $s$	$v^2 = u^2 + 2as$
final velocity $v$	$s = \tfrac{1}{2}(u+v)t$
time $t$	$v^2 = u^2 + 2as$
acceleration $a$	$s = \tfrac{1}{2}(u+v)t$
initial velocity $u$	$s = vt - \tfrac{1}{2}at^2$

1.3 Worked example: multi-stage motion

Problem. A car accelerates uniformly from rest at $2\;\mathrm{m\,s^{-2}}$ for 6 seconds, then decelerates uniformly at $3\;\mathrm{m\,s^{-2}}$ until it comes to rest. Find the total distance travelled.

Stage 1: Acceleration.

$v = u + at = 0 + 2 \times 6 = 12\;\mathrm{m\,s^{-1}}$

$s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(36) = 36\;\mathrm{m}$

Stage 2: Deceleration. Now $u = 12$, $v = 0$, $a = -3$.

$t_2 = \frac{v - u}{a} = \frac{0 - 12}{-3} = 4\;\mathrm{s}$

$s_2 = \frac{1}{2}(u + v)t_2 = \frac{1}{2}(12 + 0)(4) = 24\;\mathrm{m}$

$s_{\mathrm{total}} = 36 + 24 = 60\;\mathrm{m}$

warning

Common Pitfall When a problem has multiple stages, the final velocity of one stage becomes the initial velocity of the next. Forgetting this connection is the most frequent error in multi-stage kinematics problems.

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2. Free Fall Under Gravity

2.1 The acceleration due to gravity

Near the Earth's surface, all objects in free fall (neglecting air resistance) experience the same acceleration $g$. The standard value is:

$g \approx 9.8\;\mathrm{m\,s^{-2}} \quad (\mathrm{or}\ 9.81\;\mathrm{m\,s^{-2}}\ \mathrm{for\ greater\ precision})$

The direction of $g$ is always downward. The sign convention must be established at the start of every problem.

2.2 Sign conventions

Convention A (upward positive): Displacement upward is positive, so $a = -g$.

Convention B (downward positive): Displacement downward is positive, so $a = +g$.

Both conventions are valid, but you must be consistent throughout a single problem.

2.3 Worked example: object thrown upward

Problem. A ball is thrown vertically upward at $15\;\mathrm{m\,s^{-1}}$ from a height of $2\;\mathrm{m}$ above the ground. Taking $g = 9.8\;\mathrm{m\,s^{-2}}$ and upward as positive, find the speed with which it hits the ground.

At the highest point, $v = 0$:

$t_{\mathrm{max}} = \frac{v - u}{a} = \frac{0 - 15}{-9.8} = \frac{15}{9.8} \approx 1.531\;\mathrm{s}$

Maximum height above the throw point:

$s_{\mathrm{up}} = \frac{v^2 - u^2}{2a} = \frac{0 - 225}{2(-9.8)} = \frac{225}{19.6} \approx 11.48\;\mathrm{m}$

Total height above ground: $11.48 + 2 = 13.48\;\mathrm{m}$.

On the way down: $u = 0$, $a = -9.8$ (still upward positive), $s = -13.48\;\mathrm{m}$.

$v^2 = 0 + 2(-9.8)(-13.48) = 264.21$

$v = -\sqrt{264.21} \approx -16.26\;\mathrm{m\,s^{-1}}$

The negative sign confirms downward motion. Speed $= 16.3\;\mathrm{m\,s^{-1}}$ (3 s.f.).

info

Note that the total time of flight can also be found directly: $s = -2$, $u = 15$, $a = -9.8$: $-2 = 15t - 4.9t^2$, giving $t \approx 3.15\;\mathrm{s}$. This is not $2 \times t_{\mathrm{max}}$ because the ball was thrown from a height, not from ground level.

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3. Projectile Motion

3.1 Resolving into horizontal and vertical components

For a projectile launched with speed $u$ at angle $\theta$ above the horizontal:

$u_x = u\cos\theta, \qquad u_y = u\sin\theta$

The key principle is that horizontal and vertical motion are independent:

• Horizontal: constant velocity (no acceleration, neglecting air resistance).
• Vertical: uniform acceleration $g$ downward.

3.2 Position as a function of time

Taking the launch point as the origin, with upward as positive:

x = u\cos\theta \cdot t \tag{horizontal}

y = u\sin\theta \cdot t - \frac{1}{2}gt^2 \tag{vertical}

3.3 The trajectory equation

Eliminating $t$ from the parametric equations:

$t = \frac◆LB◆x◆RB◆◆LB◆u\cos\theta◆RB◆$

$y = u\sin\theta \cdot \frac◆LB◆x◆RB◆◆LB◆u\cos\theta◆RB◆ - \frac{1}{2}g\!\left(\frac◆LB◆x◆RB◆◆LB◆u\cos\theta◆RB◆\right)^{\!2}$

$\boxed{y = x\tan\theta - \frac◆LB◆gx^2◆RB◆◆LB◆2u^2\cos^2\theta◆RB◆}$

This is the equation of a parabola, confirming that the trajectory of a projectile (under constant gravity with no air resistance) is parabolic.

3.4 Key results

Time of flight (landing at the same height): Setting $y = 0$:

$t = \frac◆LB◆2u\sin\theta◆RB◆◆LB◆g◆RB◆$

Maximum height:

$H = \frac◆LB◆u^2\sin^2\theta◆RB◆◆LB◆2g◆RB◆$

Range:

$R = \frac◆LB◆u^2\sin 2\theta◆RB◆◆LB◆g◆RB◆$

Maximum range occurs when $\sin 2\theta = 1$, i.e. $\theta = 45^\circ$, giving $R_{\max} = \dfrac{u^2}{g}$.

3.5 Proof that complementary angles give the same range

If $\theta_1 + \theta_2 = 90^\circ$, then $\sin 2\theta_1 = \sin(180^\circ - 2\theta_2) = \sin 2\theta_2$.

Therefore $R(\theta_1) = R(\theta_2)$.

3.6 Worked example

Problem. A cricketer hits a ball at $25\;\mathrm{m\,s^{-1}}$ at $35^\circ$ above the horizontal from a height of $1.5\;\mathrm{m}$. Taking $g = 9.8\;\mathrm{m\,s^{-2}}$, find the horizontal distance travelled before the ball hits the ground.

$u_x = 25\cos 35^\circ \approx 20.48\;\mathrm{m\,s^{-1}}$

$u_y = 25\sin 35^\circ \approx 14.34\;\mathrm{m\,s^{-1}}$

When the ball hits the ground, $y = -1.5\;\mathrm{m}$:

$-1.5 = 14.34t - 4.9t^2$

$4.9t^2 - 14.34t - 1.5 = 0$

$t = \frac◆LB◆14.34 \pm \sqrt{14.34^2 + 4(4.9)(1.5)}◆RB◆◆LB◆2(4.9)◆RB◆ = \frac◆LB◆14.34 \pm \sqrt{205.64 + 29.4}◆RB◆◆LB◆9.8◆RB◆$

$t = \frac{14.34 + 15.38}{9.8} \approx 3.025\;\mathrm{s} \quad (\mathrm{taking\ the\ positive\ root})$

$x = 20.48 \times 3.025 \approx 61.95\;\mathrm{m}$

The ball travels approximately $62.0\;\mathrm{m}$ horizontally.

warning

warning When a projectile is launched from a height above the landing level, the trajectory is not symmetric. The time of ascent is less than the time of descent, and the landing angle is steeper than the launch angle.

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4. Two-Dimensional Motion with Non-Perpendicular Components

4.1 Resolving along arbitrary directions

Sometimes it is convenient to resolve velocity or acceleration along non-horizontal/vertical directions, such as parallel and perpendicular to an inclined plane.

For an inclined plane at angle $\alpha$ to the horizontal:

• Parallel to the plane: $a_{\parallel} = g\sin\alpha$ (down the plane)
• Perpendicular to the plane: $a_{\perp} = g\cos\alpha$ (into the plane)

4.2 Worked example: projectile on an inclined plane

Problem. A particle is projected up a plane inclined at $30^\circ$ to the horizontal with speed $20\;\mathrm{m\,s^{-1}}$ at an angle of $50^\circ$ to the horizontal. Taking $g = 9.8\;\mathrm{m\,s^{-1}}$, find the distance travelled up the plane before the particle lands on it.

Resolve parallel and perpendicular to the plane. The angle of projection relative to the plane is $50^\circ - 30^\circ = 20^\circ$.

Parallel to plane: $u_{\parallel} = 20\cos 20^\circ \approx 18.79\;\mathrm{m\,s^{-1}}$

$a_{\parallel} = -g\sin 30^\circ = -4.9\;\mathrm{m\,s^{-2}}$

Perpendicular to plane: $u_{\perp} = 20\sin 20^\circ \approx 6.84\;\mathrm{m\,s^{-1}}$

$a_{\perp} = -g\cos 30^\circ = -8.49\;\mathrm{m\,s^{-2}}$

Time of flight: the particle lands when its perpendicular displacement returns to zero.

$s_{\perp} = u_{\perp}t + \tfrac{1}{2}a_{\perp}t^2 = 0$

$t\!\left(6.84 - 4.245t\right) = 0 \implies t = \frac{6.84}{4.245} \approx 1.611\;\mathrm{s}$

Distance up the plane:

$s_{\parallel} = 18.79(1.611) + \tfrac{1}{2}(-4.9)(1.611)^2 = 30.27 - 6.36 \approx 23.9\;\mathrm{m}$

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5. Variable Acceleration

5.1 Using calculus for non-constant acceleration

When acceleration is not constant, the SUVAT equations do not apply. Instead, we use the calculus relationships:

$v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = v\frac{dv}{ds}$

$s = \int v\,dt, \qquad v = \int a\,dt$

The chain rule form $a = v\,\dfrac{dv}{ds}$ is particularly useful when acceleration is given as a function of displacement rather than time.

5.2 Derivation of $a = v\,\dfrac{dv}{ds}$

By the chain rule:

$a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \cdot v$

This allows us to solve problems where $a = f(s)$ by separating variables:

$v\,dv = a\,ds \implies \int v\,dv = \int a\,ds \implies \frac{1}{2}v^2 = \int a\,ds + C$

5.3 Worked example: $a = f(t)$

Problem. A particle moves in a straight line with acceleration $a = 6t - 2t^2\;\mathrm{m\,s^{-2}}$. At $t = 0$, $v = 3\;\mathrm{m\,s^{-1}}$ and $s = 0$. Find the distance travelled in the first $4$ seconds.

$v = \int (6t - 2t^2)\,dt = 3t^2 - \frac{2}{3}t^3 + C$

When $t = 0$, $v = 3$: $C = 3$.

$v = 3t^2 - \frac{2}{3}t^3 + 3$

Check if the particle changes direction (i.e. $v = 0$):

$3t^2 - \frac{2}{3}t^3 + 3 = 0 \implies 9t^2 - 2t^3 + 9 = 0$

By inspection or numerical methods, $v \gt 0$ for all $t \geq 0$ (since $9t^2 + 9 \gt 2t^3$ for $0 \leq t \leq 4$).

$s = \int_0^4 \!\left(3t^2 - \frac{2}{3}t^3 + 3\right)dt = \left[t^3 - \frac{1}{6}t^4 + 3t\right]_0^4$

$s = 64 - \frac{256}{6} + 12 = 64 - 42.67 + 12 = 33.33\;\mathrm{m}$

5.4 Worked example: $a = f(v)$

Problem. A particle moves with acceleration $a = -0.1v^2\;\mathrm{m\,s^{-2}}$. Initially $v = 10\;\mathrm{m\,s^{-1}}$. Find an expression for $v$ in terms of $t$.

Since $a = \dfrac{dv}{dt}$:

$\frac{dv}{dt} = -0.1v^2$

Separating variables:

$\int \frac{1}{v^2}\,dv = \int -0.1\,dt$

$-\frac{1}{v} = -0.1t + C$

When $t = 0$, $v = 10$: $-\dfrac{1}{10} = C$.

$-\frac{1}{v} = -0.1t - \frac{1}{10}$

$\frac{1}{v} = 0.1t + 0.1 = 0.1(t + 1)$

$v = \frac{10}{t + 1}\;\mathrm{m\,s^{-1}}$

warning

warning For variable acceleration problems, always include the constant of integration and use the initial conditions to find it. Also check whether the particle changes direction by finding when $v = 0$ -- the total distance is not the same as the displacement if there is a change of direction.

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6. Practice Problems

Problem 1

A train decelerates uniformly from $40\;\mathrm{m\,s^{-1}}$ to rest over a distance of $800\;\mathrm{m}$. Find the deceleration and the time taken.

Solution

Using $v^2 = u^2 + 2as$:

$0 = 1600 + 2a(800) \implies a = -1\;\mathrm{m\,s^{-2}}$

$t = \frac{v - u}{a} = \frac{-40}{-1} = 40\;\mathrm{s}$

Problem 2

A projectile is launched from ground level at $30\;\mathrm{m\,s^{-1}}$ at $60^\circ$ to the horizontal. Taking $g = 9.8\;\mathrm{m\,s^{-2}}$, find the maximum height and the range.

Solution

$H = \frac◆LB◆u^2\sin^2\theta◆RB◆◆LB◆2g◆RB◆ = \frac◆LB◆900 \times 0.75◆RB◆◆LB◆19.6◆RB◆ = \frac{675}{19.6} \approx 34.4\;\mathrm{m}$

$R = \frac◆LB◆u^2\sin 2\theta◆RB◆◆LB◆g◆RB◆ = \frac◆LB◆900 \times \sin 120^\circ◆RB◆◆LB◆9.8◆RB◆ = \frac◆LB◆900 \times 0.866◆RB◆◆LB◆9.8◆RB◆ \approx 79.6\;\mathrm{m}$

Problem 3

A particle moves with acceleration $a = 12t\;\mathrm{m\,s^{-2}}$. At $t = 0$, it is at rest at the origin. Find its displacement when $t = 3\;\mathrm{s}$.

Solution

$v = \int 12t\,dt = 6t^2 + C_1$

When $t = 0$, $v = 0$: $C_1 = 0$, so $v = 6t^2$.

$s = \int 6t^2\,dt = 2t^3 + C_2$

When $t = 0$, $s = 0$: $C_2 = 0$, so $s = 2t^3$.

At $t = 3$: $s = 2(27) = 54\;\mathrm{m}$.

Problem 4

A stone is thrown horizontally at $8\;\mathrm{m\,s^{-1}}$ from the top of a cliff $60\;\mathrm{m}$ high. Taking $g = 9.8\;\mathrm{m\,s^{-2}}$, find the horizontal distance from the base of the cliff where the stone lands, and the velocity (magnitude and direction) at impact.

Solution

Vertical: $s = -60$, $u_y = 0$, $a = -9.8$.

$-60 = 0 - \frac{1}{2}(9.8)t^2 \implies t = \sqrt◆LB◆\frac{120}{9.8}◆RB◆ \approx 3.50\;\mathrm{s}$

Horizontal: $x = 8 \times 3.50 = 28.0\;\mathrm{m}$.

At impact: $v_y = -9.8 \times 3.50 = -34.3\;\mathrm{m\,s^{-1}}$, $v_x = 8\;\mathrm{m\,s^{-1}}$.

$|\mathbf{v}| = \sqrt{8^2 + 34.3^2} = \sqrt{64 + 1176.5} \approx 35.2\;\mathrm{m\,s^{-1}}$

Angle below horizontal: $\theta = \arctan\!\left(\dfrac{34.3}{8}\right) \approx 76.9^\circ$.

Problem 5

A particle moves in a straight line so that its acceleration is given by $a = 4 - 2s\;\mathrm{m\,s^{-2}}$, where $s$ is the displacement from a fixed point. When $s = 0$, $v = 2\;\mathrm{m\,s^{-1}}$. Find the maximum displacement.

Solution

Using $a = v\,\dfrac{dv}{ds}$:

$v\,\frac{dv}{ds} = 4 - 2s$

Integrating:

$\int v\,dv = \int (4 - 2s)\,ds$

$\frac{1}{2}v^2 = 4s - s^2 + C$

When $s = 0$, $v = 2$: $\dfrac{1}{2}(4) = 0 + C \implies C = 2$.

$\frac{1}{2}v^2 = 4s - s^2 + 2$

Maximum displacement when $v = 0$:

$0 = 4s - s^2 + 2 \implies s^2 - 4s - 2 = 0$

$s = \frac◆LB◆4 \pm \sqrt{16 + 8}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆4 \pm \sqrt{24}◆RB◆◆LB◆2◆RB◆ = 2 \pm \sqrt{6}$

Taking the positive root: $s_{\max} = 2 + \sqrt{6} \approx 4.45\;\mathrm{m}$.