Kinematics (Extended)

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.

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:

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

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

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 $u = 12$, $v = 0$, $a = -3$.

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

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

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

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.

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.

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.).

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.

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}

Eliminating $t$ from the parametric equations:

$t = \frac{x}{u\cos\theta}$

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

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

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

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

$t = \frac{2u\sin\theta}{g}$

Maximum height:

$H = \frac{u^2\sin^2\theta}{2g}$

Range:

$R = \frac{u^2\sin 2\theta}{g}$

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

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

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

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, $y = -1.5\;\mathrm{m}$:

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

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

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

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

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

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

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)

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:

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

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.

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:

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

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 $0 \leq t \leq 4$).

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

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

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

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

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

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

• Momentum: Kinematics describes motion
• Energy and Work: Kinematic equations relate to energy
• Forces: Forces cause acceleration