Kinematics

Kinematics

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1. Fundamental Definitions

Kinematics is the mathematical description of motion, without reference to the forces that cause it. We begin with three rigorous definitions.

Definition. The displacement $s$ of a particle is its position vector relative to a chosen origin. Unlike distance, displacement is a vector quantity.

Definition. The velocity $v$ of a particle is the rate of change of its displacement with respect to time:

$v = \frac{ds}{dt} = \lim_{\Delta t \to 0}\frac◆LB◆\Delta s◆RB◆◆LB◆\Delta t◆RB◆$

Velocity is a vector. Its magnitude is the speed.

Definition. The acceleration $a$ of a particle is the rate of change of its velocity with respect to time:

$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

Acceleration is also a vector. The SI unit is m s$^{-2}$.

2. Derivation of the SUVAT Equations

For uniform acceleration (constant $a$), we derive five equations relating the kinematic variables $s$, $u$, $v$, $a$, and $t$.

Equation 1: $v = u + at$

Starting from the definition of acceleration:

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

Since $a$ is constant, we integrate with respect to time:

$\int_{u}^{v} dv = \int_{0}^{t} a \, dt' \implies v - u = at \implies \boxed{v = u + at}$

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

From the definition of velocity:

$v = \frac{ds}{dt}$

For constant acceleration, the velocity varies linearly from $u$ to $v$. The average velocity over the interval is $\bar{v} = \frac{u + v}{2}$. Since displacement equals average velocity multiplied by time:

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

Rigorous derivation. Integrating $ds = v \, dt$ and substituting $v = u + at$:

$s = \int_0^t (u + at')\,dt' = ut + \frac{1}{2}at^2$

From Equation 1, $v = u + at$, so $at = v - u$, giving:

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

Equation 3: $s = ut + \frac{1}{2}at^2$

This was obtained above during the rigorous derivation of Equation 2:

$\boxed{s = ut + \frac{1}{2}at^2}$

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

We eliminate $t$ between Equations 1 and 3. From Equation 1: $t = \frac{v - u}{a}$. Substituting into Equation 3:

$$\begin{aligned} s &= u\left(\frac{v - u}{a}\right) + \frac{1}{2}a\left(\frac{v - u}{a}\right)^2 \\[4pt] s &= \frac{u(v - u)}{a} + \frac{(v - u)^2}{2a} \\[4pt] 2as &= 2u(v - u) + (v - u)^2 \\[4pt] 2as &= 2uv - 2u^2 + v^2 - 2uv + u^2 \\[4pt] 2as &= v^2 - u^2 \end{aligned}$$

$\boxed{v^2 = u^2 + 2as}$

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

Substitute $u = v - at$ (from Equation 1) into Equation 3:

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

$\boxed{s = vt - \frac{1}{2}at^2}$

tip

Exam Technique To decide which SUVAT equation to use, identify which variable is not given and not asked for, then use the equation that doesn't contain it.

Missing variable	Use
$s$	$v = u + at$
$v$	$s = ut + \frac{1}{2}at^2$
$u$	$s = vt - \frac{1}{2}at^2$
$a$	$s = \frac{1}{2}(u + v)t$
$t$	$v^2 = u^2 + 2as$

3. Displacement-Time and Velocity-Time Graphs

• The gradient of a displacement-time graph gives the velocity.
• The gradient of a velocity-time graph gives the acceleration.
• The area under a velocity-time graph gives the displacement.

The Moving Man

Explore the simulation above to develop intuition for this topic.

For uniform acceleration, the $v$-$t$ graph is a straight line, and the area is a trapezium.

For non-constant acceleration, the $v$-$t$ graph is curved, but the same three principles apply at every instant:

• The gradient of the $s$-$t$ curve at any point gives the instantaneous velocity at that time.
• The gradient of the $v$-$t$ curve at any point gives the instantaneous acceleration at that time.
• The area under the $v$-$t$ curve between two times gives the displacement over that interval.

When the $v$-$t$ graph is curved, the area cannot be found using the trapezium rule for a single straight line. Two approaches are available:

1. Counting squares -- estimate the area by counting grid squares on the graph paper, treating partial squares by eye.
2. Integration -- if $v(t)$ is given algebraically, compute $\int_{t_1}^{t_2} v(t)\,dt$ exactly.

tip

Exam Technique When asked to find the distance travelled from a curved $v$-$t$ graph, check whether the curve crosses the time axis. If it does, the velocity changes sign and you must split the calculation: the area above the axis is positive displacement, the area below is negative displacement. The total distance is the sum of absolute areas; the net displacement is their algebraic sum.

warning

Common Pitfall Students often draw a tangent to a curve incorrectly by placing the ruler away from the point of interest. Always place the ruler so it just touches the curve at the point where you need the gradient, then extend it to read two clear coordinates for $\Delta y / \Delta x$.

Example: Finding displacement from a curved v-t graph

A particle has velocity $v = 6t - t^2$ m s$^{-1}$ for $0 \leq t \leq 6$ s. The $v$-$t$ graph is a parabola opening downward, with $v = 0$ at $t = 0$ and $t = 6$, and maximum $v = 9$ m s$^{-1}$ at $t = 3$. The displacement over the full 6 seconds is $\int_0^6 (6t - t^2)\,dt = [3t^2 - t^3/3]_0^6 = 108 - 72 = 36$ m.

4. Free Fall

A body in free fall moves under the influence of gravity alone (neglecting air resistance). Near the Earth's surface, all objects experience the same gravitational acceleration:

$g \approx 9.81 \mathrm{ m s}^{-2}$

This was established by Galileo's experiments and is a consequence of the equivalence principle (mass cancels in $F = ma = mg$).

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info problems. Always use the value specified in the question.

5. Projectile Motion

Assumptions

We assume:

1. The only force acting is gravity (no air resistance).
2. $g$ is constant (valid for trajectories small compared to Earth's radius).
3. The horizontal and vertical components of motion are independent.

Independence of Components

This is the central insight. Since gravity acts vertically, it produces no horizontal acceleration. Therefore:

• Horizontal: $a_x = 0 \implies v_x = \mathrm{const} = v_0\cos\theta$
• Vertical: $a_y = -g \implies$ the motion is uniformly accelerated

Deriving the Parabolic Trajectory

A projectile is launched from the origin with speed $v_0$ at angle $\theta$ above the horizontal.

Horizontal motion (constant velocity):

$x = v_0\cos\theta \cdot t \implies t = \frac◆LB◆x◆RB◆◆LB◆v_0\cos\theta◆RB◆$

Vertical motion (uniform acceleration):

$y = v_0\sin\theta \cdot t - \frac{1}{2}gt^2$

Substituting $t$ from the horizontal equation:

$$\begin{aligned} y &= v_0\sin\theta \left(\frac◆LB◆x◆RB◆◆LB◆v_0\cos\theta◆RB◆\right) - \frac{1}{2}g\left(\frac◆LB◆x◆RB◆◆LB◆v_0\cos\theta◆RB◆\right)^2 \\[4pt] y &= x\tan\theta - \frac◆LB◆gx^2◆RB◆◆LB◆2v_0^2\cos^2\theta◆RB◆ \end{aligned}$$

This is the equation of a parabola — a direct consequence of constant horizontal velocity combined with constant vertical acceleration.

Maximum Height

At maximum height, the vertical velocity is zero: $v_y = v_0\sin\theta - gt = 0$, giving $t_{\mathrm{peak}} = \frac◆LB◆v_0\sin\theta◆RB◆◆LB◆g◆RB◆$.

$H = v_0\sin\theta \cdot \frac◆LB◆v_0\sin\theta◆RB◆◆LB◆g◆RB◆ - \frac{1}{2}g\left(\frac◆LB◆v_0\sin\theta◆RB◆◆LB◆g◆RB◆\right)^2 = \frac◆LB◆v_0^2\sin^2\theta◆RB◆◆LB◆2g◆RB◆$

$\boxed{H = \frac◆LB◆v_0^2\sin^2\theta◆RB◆◆LB◆2g◆RB◆}$

Time of Flight

The projectile returns to $y = 0$ when:

$0 = v_0\sin\theta \cdot t - \frac{1}{2}gt^2 \implies t(v_0\sin\theta - \frac{1}{2}gt) = 0$

The non-trivial solution gives:

$\boxed{T = \frac◆LB◆2v_0\sin\theta◆RB◆◆LB◆g◆RB◆}$

Maximum Range

The range is $R = v_0\cos\theta \cdot T$:

$R = v_0\cos\theta \cdot \frac◆LB◆2v_0\sin\theta◆RB◆◆LB◆g◆RB◆ = \frac◆LB◆v_0^2 \cdot 2\sin\theta\cos\theta◆RB◆◆LB◆g◆RB◆ = \frac◆LB◆v_0^2\sin 2\theta◆RB◆◆LB◆g◆RB◆$

$\boxed{R = \frac◆LB◆v_0^2\sin 2\theta◆RB◆◆LB◆g◆RB◆}$

Maximum range occurs when $\sin 2\theta = 1$, i.e., $\theta = 45^\circ$:

$R_{\max} = \frac{v_0^2}{g}$

Intuition: Why Parabolas?

A parabola arises because the vertical position depends quadratically on time ($y \propto t^2$), while the horizontal position depends linearly on time ($x \propto t$). Eliminating $t$ between a linear and quadratic relation always produces a parabola. The shape is not special to gravity — it is the geometry of combining uniform motion in one direction with uniformly accelerated motion in the perpendicular direction.

Effect of Air Resistance on Projectiles

In reality, air resistance (drag) opposes the velocity of a moving object. The magnitude of the drag force depends on the object's speed and shape, and typically increases with speed.

Effect on projectile trajectory. When drag is included:

• The horizontal velocity is no longer constant -- it decreases throughout the flight because drag has a horizontal component opposing the motion.
• On the way up, both gravity and the vertical component of drag act downward, so the vertical deceleration is greater than $g$.
• On the way down, gravity acts downward but drag acts upward (opposing the downward velocity), so the vertical acceleration is less than $g$.
• The trajectory is no longer parabolic. The descent is steeper than the ascent, the range is shorter, and the maximum height is lower compared to the idealised case.

Terminal velocity. For an object falling vertically under gravity with air resistance, the drag force increases with speed. Eventually, the drag force equals the weight of the object:

$F_{\mathrm{drag}} = mg$

At this point the net force is zero, the acceleration is zero, and the object falls at a constant speed called the terminal velocity $v_T$:

$\boxed{v_T \mathrm{ is reached when } F_{\mathrm{drag}} = mg \implies a = 0}$

The $v$-$t$ graph for a falling object reaching terminal velocity shows the velocity increasing with a decreasing gradient (decreasing acceleration) until it asymptotically approaches $v_T$.

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warning

Common Pitfall Terminal velocity does not mean the object has stopped accelerating because it has run out of force. The forces are balanced: weight down equals drag up. The acceleration is zero because the net force is zero, not because no forces act.

6. Non-Uniform Acceleration

When acceleration is not constant, the SUVAT equations do not apply. Instead, we use calculus to relate displacement, velocity, and acceleration.

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The Differential Relations

From the definitions:

$\boxed{v = \frac{ds}{dt}}$

$\boxed{a = \frac{dv}{dt} = \frac{d^2s}{dt^2}}$

A third relation follows from the chain rule. Since $a = dv/dt$ and $v = ds/dt$:

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

$\boxed{a = v\,\frac{dv}{ds}}$

This form is useful when acceleration is given as a function of displacement $s$ rather than time $t$.

Selecting the Correct Form

Given	Use
$a = f(t)$	$a = dv/dt$, integrate with respect to $t$
$a = f(v)$	Rewrite as $dt/dv = 1/f(v)$ and integrate, or use $a = v\,dv/ds$
$a = f(s)$	$a = v\,dv/ds$, integrate with respect to $s$

tip

Exam Technique Identify what the acceleration is given as a function of -- time, velocity, or displacement -- to decide which calculus method to use. If $a$ is a function of $t$, integrate $dv/dt$ directly. If $a$ is a function of $s$, use $a = v\,dv/ds$ and integrate with respect to $s$. If $a$ is a function of $v$, rewrite as $dt/dv = 1/f(v)$ and integrate to find $t(v)$, then invert if possible.

Derivation of $v^2 = u^2 + 2as$ by Integration

Starting from $a = v\,dv/ds$ and assuming constant $a$:

$\int_{u}^{v} v'\,dv' = \int_{0}^{s} a\,ds'$

$\left[\frac{v'^2}{2}\right]_{u}^{v} = as$

$\frac{v^2}{2} - \frac{u^2}{2} = as$

$\boxed{v^2 = u^2 + 2as} \quad \square$

This derivation relies only on the chain-rule identity $a = v\,dv/ds$.

Example: Variable acceleration

A particle moves in a straight line with acceleration $a = 6t$ m s$^{-2}$, starting from rest at the origin.

(a) Find $v$ as a function of $t$. $a = dv/dt = 6t$, so $dv = 6t\,dt$. Integrating: $v = \int_0^t 6t'\,dt' = 3t^2 + C$. Since $v(0) = 0$: $C = 0$, giving $v = 3t^2$ m s$^{-1}$.

(b) Find $s$ as a function of $t$. $v = ds/dt = 3t^2$, so $ds = 3t^2\,dt$. Integrating: $s = \int_0^t 3t'^2\,dt' = t^3$ m.

(c) Find the acceleration as a function of $s$. From $v = 3t^2$ and $s = t^3$: $t = s^{1/3}$, so $v = 3s^{2/3}$. Then $a = v\,dv/ds = 3s^{2/3} \cdot 2s^{-1/3} = 6s^{1/3} = 6t$, confirming consistency.

7. Relative Velocity in Two Dimensions

When two objects move in different directions, their velocities are combined using vector addition. The velocity of A relative to B is:

$\boxed{\mathbf{v}_{AB} = \mathbf{v}_A - \mathbf{v}_B}$

This is the velocity A appears to have when observed from B's frame of reference.

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Resolving into Components

In two dimensions, resolve each velocity into perpendicular components, then subtract component by component.

Example. A river flows east at $3.0$ m s$^{-1}$. A boat heads north at $5.0$ m s$^{-1}$ relative to the water. The boat's velocity relative to the ground is:

$\mathbf{v}_{\mathrm{ground}} = (3.0\ \mathrm{east}) + (5.0\ \mathrm{north})$

Speed relative to ground: $v = \sqrt{3^2 + 5^2} = \sqrt{34} = 5.83$ m s$^{-1}$.

Direction: $\theta = \arctan(5/3) = 59.0^\circ$ north of east.

Crossing a River

A classic relative-velocity problem: to cross a river of width $d$ with current speed $v_c$, a boat with speed $v_b$ (relative to water) can:

• Head directly across (perpendicular to bank). The boat is carried downstream. Time to cross: $t = d/v_b$. Downstream displacement: $\Delta x = v_c \cdot d/v_b$.
• Head upstream at an angle to compensate for the current. If the boat heads at angle $\theta$ upstream from the perpendicular, the component across the river is $v_b\cos\theta$ and the component against the current is $v_b\sin\theta$. For zero downstream drift: $v_b\sin\theta = v_c$, giving $\theta = \arcsin(v_c/v_b)$. This is only possible if $v_b > v_c$.

warning

Common Pitfall The boat's speed $v_b$ is always relative to the water, not relative to the ground. The ground speed is the vector sum of the boat's water-relative velocity and the current velocity. Never add the magnitudes directly unless the velocities are in the same direction.

Relative Velocity and Closest Approach

Two objects A and B are on a collision course if their relative velocity $\mathbf{v}_{AB}$ is directed along the line joining them. If not, the closest approach occurs when the position vector from A to B is perpendicular to the relative velocity: $\mathbf{r}_{BA} \cdot \mathbf{v}_{BA} = 0$.

Example: Will two ships collide?

Ship A sails north at $10$ m s$^{-1}$ and is at position $(0, 0)$ m. Ship B sails east at $10$ m s$^{-1}$ and is at position $(200, 0)$ m. Will they collide?

$\mathbf{v}_{BA} = \mathbf{v}_B - \mathbf{v}_A = (10\ \mathrm{east}) - (10\ \mathrm{north}) = (10, -10)$ m s$^{-1}$.

The position of B relative to A is $(200, 0)$. The relative velocity $(10, -10)$ is not parallel to $(200, 0)$, so they will not collide. The closest approach occurs when $\mathbf{r}_{BA} \cdot \mathbf{v}_{BA} = 0$.

Problem Set

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Problem 1

A stone is thrown vertically upward with speed $15.0$ m s$^{-1}$. Find: (a) the maximum height reached, (b) the time to reach maximum height, (c) the total time of flight, (d) the speed when it returns to the thrower's hand.

Answer. (a) $v^2 = u^2 + 2as$: $0 = 15^2 - 2(9.81)s$, $s = \frac{225}{19.62} = 11.5$ m.

(b) $v = u + at$: $0 = 15 - 9.81t$, $t = 1.53$ s.

(c) By symmetry (no air resistance), $T = 2t = 3.06$ s.

(d) By conservation of energy (or symmetry), the speed equals the initial speed: $15.0$ m s$^{-1}$.

If you get this wrong, revise: Free Fall and SUVAT Equations

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Problem 2

A projectile is launched from ground level with speed $25.0$ m s$^{-1}$ at an angle of $35^\circ$ to the horizontal. Calculate the horizontal range and the maximum height.

Answer. $R = \frac◆LB◆v_0^2\sin 2\theta◆RB◆◆LB◆g◆RB◆ = \frac◆LB◆625 \times \sin 70°◆RB◆◆LB◆9.81◆RB◆ = \frac◆LB◆625 \times 0.9397◆RB◆◆LB◆9.81◆RB◆ = 59.9$ m.

$H = \frac◆LB◆v_0^2\sin^2\theta◆RB◆◆LB◆2g◆RB◆ = \frac◆LB◆625 \times \sin^2 35°◆RB◆◆LB◆19.62◆RB◆ = \frac◆LB◆625 \times 0.329◆RB◆◆LB◆19.62◆RB◆ = 10.5$ m.

If you get this wrong, revise: Maximum Height and Maximum Range

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Problem 3

A ball is thrown from a cliff of height $40.0$ m with horizontal velocity $12.0$ m s$^{-1}$. Find: (a) the time to hit the ground, (b) the horizontal distance from the cliff base, (c) the vertical component of velocity at impact, (d) the magnitude of the final velocity.

Answer. (a) Vertical: $s = ut + \frac{1}{2}at^2$, $40 = 0 + \frac{1}{2}(9.81)t^2$, $t = \sqrt◆LB◆\frac{80}{9.81}◆RB◆ = 2.86$ s.

(b) $R = v_x t = 12.0 \times 2.86 = 34.3$ m.

(c) $v_y = 0 + 9.81 \times 2.86 = 28.1$ m s$^{-1}$ downward.

(d) $v = \sqrt{12^2 + 28.1^2} = \sqrt{144 + 789.6} = \sqrt{933.6} = 30.6$ m s$^{-1}$.

If you get this wrong, revise: Independence of Components

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Problem 4

Derive the equation $s = ut + \frac{1}{2}at^2$ from the definition of velocity as a derivative, assuming constant acceleration.

Answer. $v = \frac{ds}{dt}$ and $a = \frac{dv}{dt} = \mathrm{const}$. Integrating: $\int_0^t a\,dt' = \int_u^v dv'$, giving $v = u + at$. Then $ds/dt = u + at$. Integrating: $\int_0^s ds' = \int_0^t (u + at')\,dt' = [ut' + \frac{1}{2}at'^2]_0^t = ut + \frac{1}{2}at^2$. $\square$

If you get this wrong, revise: Derivation of the SUVAT Equations

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Problem 5

Two balls are thrown simultaneously from the same height. Ball A is thrown vertically upward with speed $20$ m s$^{-1}$; Ball B is thrown vertically downward with speed $20$ m s$^{-1}$. Which ball hits the ground first, and by how much? Take the initial height as $30$ m.

Answer. For Ball A: upward first, then downward. Time to reach max height: $t_1 = 20/9.81 = 2.04$ s, max height above launch: $20^2/(2 \times 9.81) = 20.4$ m, total height above ground = $50.4$ m. Time to fall from 50.4 m: $50.4 = \frac{1}{2}(9.81)t_2^2$, $t_2 = 3.21$ s. Total: $t_A = 2.04 + 3.21 = 5.25$ s.

For Ball B: $30 = 20t + \frac{1}{2}(9.81)t^2$. Solving: $4.905t^2 + 20t - 30 = 0$, $t = \frac◆LB◆-20 + \sqrt{400 + 588.6}◆RB◆◆LB◆9.81◆RB◆ = \frac{-20 + 31.4}{9.81} = 1.16$ s.

Ball B hits first. Difference: $5.25 - 1.16 = 4.09$ s.

If you get this wrong, revise: SUVAT Equations

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Problem 6

A particle moves along a straight line with velocity $v = 3t^2 - 6t$ m s$^{-1}$ for $0 \leq t \leq 4$ s. Find: (a) when the particle is at rest, (b) the total distance travelled, (c) the displacement.

Answer. (a) $v = 0$: $3t^2 - 6t = 0 \implies t = 0$ or $t = 2$ s.

(b) $s(t) = \int v\,dt = t^3 - 3t^2 + C$. Take $s(0) = 0$, so $C = 0$.

• At $t = 2$: $s = 8 - 12 = -4$ m.
• At $t = 4$: $s = 64 - 48 = 16$ m. Total distance = $4 + 20 = 24$ m.

(c) Displacement = $16$ m.

If you get this wrong, revise: Displacement-Time and Velocity-Time Graphs

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Problem 7

A projectile is fired with speed $v_0$ at angle $\theta$ from a height $h$ above level ground. Derive an expression for the time of flight in terms of $v_0$, $\theta$, $g$, and $h$.

Answer. $y = h + v_0\sin\theta \cdot t - \frac{1}{2}gt^2 = 0$ at landing. This gives $\frac{1}{2}gt^2 - v_0\sin\theta \cdot t - h = 0$. By the quadratic formula:

$t = \frac◆LB◆v_0\sin\theta + \sqrt◆LB◆v_0^2\sin^2\theta + 2gh◆RB◆◆RB◆◆LB◆g◆RB◆$

(We take the positive root since $t > 0$.)

If you get this wrong, revise: Deriving the Parabolic Trajectory

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Problem 8

A car accelerates uniformly from $10$ m s$^{-1}$ to $30$ m s$^{-1}$ while covering a distance of $200$ m. Find the acceleration and the time taken.

Answer. Using $v^2 = u^2 + 2as$: $900 = 100 + 2a(200)$, $400a = 800$, $a = 2.0$ m s$^{-2}$.

Using $v = u + at$: $30 = 10 + 2t$, $t = 10$ s.

If you get this wrong, revise: SUVAT Equations

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Problem 9

A golfer hits a ball from the ground with speed $40$ m s$^{-1}$. At what angle should she hit the ball to land it $120$ m away? (Give both possible angles.)

Answer. $R = \frac◆LB◆v_0^2\sin 2\theta◆RB◆◆LB◆g◆RB◆$, so $\sin 2\theta = \frac{Rg}{v_0^2} = \frac◆LB◆120 \times 9.81◆RB◆◆LB◆1600◆RB◆ = 0.736$.

$2\theta = 47.4^\circ$ or $132.6^\circ$, giving $\theta = 23.7^\circ$ or $66.3^\circ$. Both angles give the same range — complementary angles always do (since $\sin 2\theta = \sin(180° - 2\theta)$).

If you get this wrong, revise: Maximum Range

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Problem 10

On the Moon, $g = 1.62$ m s$^{-2}$. A astronaut throws a rock with speed $12$ m s$^{-1}$ at $60^\circ$ to the horizontal. Compare the maximum height and range to what they would be on Earth.

Answer. On the Moon: $H = \frac◆LB◆144 \times \sin^2 60°◆RB◆◆LB◆2 \times 1.62◆RB◆ = \frac◆LB◆144 \times 0.75◆RB◆◆LB◆3.24◆RB◆ = 33.3$ m. $R = \frac◆LB◆144 \times \sin 120°◆RB◆◆LB◆1.62◆RB◆ = \frac◆LB◆144 \times 0.866◆RB◆◆LB◆1.62◆RB◆ = 77.0$ m.

On Earth: $H = \frac{108}{19.62} = 5.51$ m. $R = \frac{124.7}{9.81} = 12.7$ m.

The Moon gives $\frac{33.3}{5.51} = 6.0$ times greater height and $\frac{77.0}{12.7} = 6.1$ times greater range, consistent with the ratio $g_E/g_M \approx 6.1$.

If you get this wrong, revise: Maximum Height and Maximum Range

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Problem 11

A ball rolls off a table of height $1.2$ m with a horizontal speed of $3.0$ m s$^{-1}$. A second ball is dropped from the same height at the same instant. Which ball hits the ground first? Justify your answer.

Answer. Both balls hit the ground at the same time. The horizontal velocity of the first ball does not affect its vertical motion. Both have $u_y = 0$, $a_y = g$, and $h = 1.2$ m, so both have $t = \sqrt{2h/g} = 0.495$ s.

If you get this wrong, revise: Independence of Components

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Problem 12

An object moves with uniform acceleration. In the first 3 seconds it travels 18 m, and in the next 2 seconds it travels 30 m. Find the initial velocity and the acceleration.

Answer. Using $s = ut + \frac{1}{2}at^2$:

• For $t = 3$: $18 = 3u + \frac{9}{2}a$ ... (i)
• For $t = 5$: $18 + 30 = 5u + \frac{25}{2}a$ ... (ii), i.e., $48 = 5u + \frac{25}{2}a$.

From (i): $3u = 18 - 4.5a$, $u = 6 - 1.5a$. Substituting into (ii): $48 = 5(6 - 1.5a) + 12.5a = 30 - 7.5a + 12.5a = 30 + 5a$. $5a = 18$, $a = 3.6$ m s$^{-2}$. $u = 6 - 1.5(3.6) = 6 - 5.4 = 0.6$ m s$^{-1}$.

If you get this wrong, revise: SUVAT Equations

Problem 13

A particle moves along a straight line with acceleration $a = 4 - 2t$ m s$^{-2}$. At $t = 0$, the particle is at rest at the origin. Find: (a) the velocity as a function of $t$, (b) the displacement as a function of $t$, (c) the time at which the particle is momentarily at rest again, (d) the displacement at that time.

Answer. (a) $a = dv/dt = 4 - 2t$. Integrating: $v = 4t - t^2$ m s$^{-1}$ (using $v(0) = 0$).

(b) $ds/dt = 4t - t^2$. Integrating: $s = 2t^2 - t^3/3$ m (using $s(0) = 0$).

(c) $v = 0$: $t(4 - t) = 0 \implies t = 4$ s.

(d) $s(4) = 2(16) - 64/3 = 32 - 21.3 = 10.7$ m.

If you get this wrong, revise: Non-Uniform Acceleration

Problem 14

A river is $80$ m wide and flows at $2.5$ m s$^{-1}$. A boat can travel at $5.0$ m s$^{-1}$ in still water. (a) If the boat heads directly across the river, how long does it take to cross and how far downstream does it land? (b) At what angle upstream should the boat head to land directly opposite the starting point?

Answer. (a) Time = $80/5.0 = 16$ s. Downstream displacement = $2.5 \times 16 = 40$ m.

(b) For zero downstream drift: $\sin\theta = 2.5/5.0 = 0.5$, giving $\theta = 30^\circ$ upstream from the perpendicular.

If you get this wrong, revise: Crossing a River

Problem 15

A particle moves with acceleration $a = -0.5v$ m s$^{-2}$, where $v$ is the speed. The initial speed is $10$ m s$^{-1}$. Find: (a) the velocity as a function of time, (b) the time taken for the speed to halve.

Answer. (a) $dv/dt = -0.5v$. Separating variables: $\frac{dv}{v} = -0.5\,dt$. Integrating: $\ln v = -0.5t + C$. At $t = 0$, $v = 10$: $\ln 10 = C$. So $v = 10e^{-0.5t}$ m s$^{-1}$.

(b) When $v = 5$: $5 = 10e^{-0.5t}$, $e^{-0.5t} = 0.5$, $t = 1.39$ s.

If you get this wrong, revise: Non-Uniform Acceleration

Problem 16

A particle moves with acceleration $a = 3s$ m s$^{-2}$, where $s$ is the displacement from the origin. At $s = 0$, the velocity is $4.0$ m s$^{-1}$. Find the velocity when $s = 2.0$ m.

Answer. We are given $a = f(s)$, so use $a = v\,dv/ds$:

$\int_{4}^{v} v'\,dv' = \int_{0}^{2} 3s'\,ds'$

$\frac{v^2}{2} - 8 = 6$

$v^2 = 28$, $v = 5.29$ m s$^{-1}$.

If you get this wrong, revise: The Differential Relations

Problem 17

A ball is thrown from the top of a building of height $45$ m with speed $20$ m s$^{-1}$ at an angle of $30^\circ$ above the horizontal. Find: (a) the time of flight, (b) the horizontal distance from the base where the ball lands, (c) the speed at impact.

Answer. (a) Vertical: $y = 45 + 10t - 4.905t^2 = 0$.

$4.905t^2 - 10t - 45 = 0$, $t = \frac◆LB◆10 + \sqrt{100 + 882.9}◆RB◆◆LB◆9.81◆RB◆ = 4.22$ s.

(b) $R = 20\cos 30° \times 4.22 = 17.3 \times 4.22 = 73.1$ m.

(c) $v_x = 17.3$ m s$^{-1}$, $v_y = 10 - 9.81 \times 4.22 = -31.4$ m s$^{-1}$. Speed = $\sqrt{17.3^2 + 31.4^2} = 35.8$ m s$^{-1}$.

If you get this wrong, revise: Projectile Motion

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tip

tip Ready to test your understanding of Kinematics? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Kinematics with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

danger

danger

• Applying SUVAT equations to non-uniform acceleration: The five SUVAT equations ONLY apply when acceleration is CONSTANT. If a question involves a changing force (e.g., a spring, air resistance), acceleration is not constant and SUVAT cannot be used. Use calculus (integration) or energy methods instead.

• Confusing displacement with distance: Displacement is a VECTOR (includes direction) and can be negative. Distance is a SCALAR and is always positive. In vertical motion, upward displacement is positive and downward is negative. A ball thrown up and caught at the same point has zero total displacement but non-zero distance travelled.

• Taking g as positive in both directions: When using SUVAT with vertical motion, be consistent with sign convention. If upward is positive, then g = -9.81 m/s squared (acceleration is downward). Substituting g = +9.81 with upward-positive convention will give the wrong answer for velocity and time.

• Forgetting that air resistance changes the motion: In many projectile questions, air resistance is neglected. But if a question mentions air resistance, remember: the horizontal component of velocity DECREASES (it is no longer constant), the time of flight DECREASES (the object reaches a lower maximum height), and the trajectory is no longer a perfect parabola.

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