Kinematics

Task 1 - Rates of change

Hook

A speed camera works out your speed by timing you between two lines painted on the road. A speedometer shows your speed at this exact instant. How can a speed exist at a single instant, when speed is distance divided by time and no time passes in an instant?

Intuition

Start with an average over an interval, the slope of the straight line joining two points on a position-time graph. This line is called a secant. Now shrink the interval. The second point slides toward the first, and the secant swings round until it just touches the curve at a single point. That touching line is the tangent, and its slope is the instantaneous velocity.

Shrink the interval below and watch the two readings converge.

Formalise

The derivative is the limit of the average rate of change as the interval shrinks to zero. Velocity is the derivative of position, and acceleration is the derivative of velocity, which makes it the second derivative of position. Going the other way, the area under a velocity-time graph, the integral, gives displacement.

Worked example

The curve in the diagram is $x(t) = 0.4t^2 + 0.5t$. At $t = 3\text{ s}$, compare the average velocity over the next 2.5 s, 0.1 s and 0.01 s.

1. Over 2.5 s. $x(5.5) - x(3) = 14.85 - 5.1 = 9.75$, so $\bar{v} = 9.75 / 2.5 = 3.9\text{ m/s}$.
2. Over 0.1 s. $x(3.1) - x(3) = 5.394 - 5.1 = 0.294$, so $\bar{v} = 2.94\text{ m/s}$.
3. Over 0.01 s. $\bar{v} = 2.904\text{ m/s}$.

The averages settle on 2.9 m/s, the instantaneous velocity at 3 s.

Your turn

Task 2 - The constant acceleration equations

Hook

A car brakes from 25 m/s with a constant deceleration. How far does it travel before it stops? You know the start speed, the end speed and the acceleration, but not the time. Is there an equation that skips the time completely?

Intuition

With constant acceleration, a velocity-time graph is a straight line from the initial velocity $u$ to the final velocity $v$ over a time $t$. Its gradient is $a$. The area underneath, a trapezium, is the displacement $s$. Writing down the gradient and the area in different ways gives five equations linking the five quantities $s, u, v, a, t$.

Formalise

The gradient of the line gives the first equation, and the area of the trapezium gives the second. Substituting one into the other produces the rest.

Worked example 1

The braking car. $u = 25\text{ m/s}$, $v = 0$, $a = -5\text{ m/s}^2$. Find the stopping distance and the time.

1. Distance, which needs no time. $0 = 25^2 + 2(-5)s$, so $s = 625 / 10 = 62.5\text{ m}$.
2. Time. $0 = 25 + (-5)t$, so $t = 5\text{ s}$.

Worked example 2

A ball is dropped from rest 20 m above the ground. Take down as positive, so $a = +9.81\text{ m/s}^2$.

1. Time. $20 = 0 + \tfrac{1}{2}(9.81)t^2$, so $t = \sqrt{40 / 9.81} = 2.02\text{ s}$.
2. Landing speed. $v^2 = 0 + 2(9.81)(20) = 392.4$, so $v = 19.8\text{ m/s}$.

Your turn

Task 3 - Vectors in two dimensions

Hook

A boat points straight across a river and motors at 4 m/s. The river flows sideways at 3 m/s. How fast is the boat actually moving, and in which direction?

Intuition

Any vector in a flat plane can be split into two perpendicular components, one along x and one along y. The vector is the diagonal of the rectangle they form. Splitting a vector is called resolving it, and it is what makes two-dimensional problems manageable. To add two vectors, either place them tip to tail, or simply add their x components and their y components separately.

Drag the tip of vector A below. Then switch to "Add a second vector" and drag the tip of B.

Formalise

For a vector of size $v$ at angle $\theta$ above the positive x axis.

Worked example

The boat. Take x as downstream and y as straight across.

1. Components. $v_x = 3\text{ m/s}$, $v_y = 4\text{ m/s}$.
2. Size. $\sqrt{3^2 + 4^2} = 5\text{ m/s}$.
3. Direction. $\tan^{-1}(4/3) = 53.1^\circ$ from the bank, which is $36.9^\circ$ downstream of straight across.

Your turn

Give angles in degrees from the positive x axis.

Task 4 - Projectile motion

Hook

At the same moment, one ball is fired horizontally from a table and another is simply dropped from the same height. Which lands first?

Intuition

They land together. Gravity acts only vertically, so it changes only the vertical velocity. Both balls start with zero vertical velocity and gain it at the same rate, so they fall the same distance in the same time. The fired ball also moves sideways, at a constant horizontal velocity, but that has no effect on how quickly it falls.

Fire the launcher below. The teal arrows show the horizontal velocity at regular times, and they never change. The red arrows show the vertical velocity, which shrinks to zero at the top and then grows downward. Try angles of 30° and 60°, and then find the angle that gives the longest range.

Formalise

Split the launch velocity $u$ at angle $\theta$ into $u_x = u\cos\theta$ and $u_y = u\sin\theta$. Horizontally there is no acceleration. Vertically the acceleration is $-g$. For a launch and landing at the same height, setting $y = 0$ gives the flight time, and the other results follow.

Worked example

A ball is launched at 20 m/s at 30° above level ground.

1. Components. $u_x = 20\cos 30^\circ = 17.32\text{ m/s}$, $u_y = 20\sin 30^\circ = 10\text{ m/s}$.
2. Flight time. $T = 2 \times 10 / 9.81 = 2.04\text{ s}$.
3. Range. $R = 17.32 \times 2.04 = 35.3\text{ m}$.
4. Maximum height. $H = 10^2 / (2 \times 9.81) = 5.10\text{ m}$.

Because $\sin 2\theta$ is largest at $2\theta = 90^\circ$, the range on level ground is greatest at 45°. Angles that add to 90°, such as 30° and 60°, give the same range.

Your turn

Task 5 - Designing a jump

Hook

A level designer wants the player character to jump exactly 3 m high, the height of three tiles, and to reach the top of the jump in 0.4 s so it feels snappy. What value of gravity, and what launch speed, does the game need?

Intuition

This is the SUVAT problem run backwards. Designers choose how high and how fast the jump should feel, and the physics then tells them what gravity and launch velocity to put in the code. At the top of the jump the vertical velocity is zero, which gives the extra piece of information needed.

Formalise

Taking up as positive, with $v = 0$ at the apex, $s = h$ and $a = -g$, the equations $v = u + at$ and $s = vt - \tfrac{1}{2}at^2$ give the following.

Worked example

With $h = 3\text{ m}$ and $t_{apex} = 0.4\text{ s}$.

1. Gravity. $g = 2 \times 3 / 0.4^2 = 37.5\text{ m/s}^2$.
2. Launch speed. $v_0 = 2 \times 3 / 0.4 = 15\text{ m/s}$.

That gravity is almost four times Earth's. Real people jumping 3 m would hang in the air for much longer, which feels floaty and slow in a game. Many platformers go further and use a larger gravity on the way down than on the way up, and cut the upward velocity when the jump button is released, so a tap gives a short hop and a hold gives a full jump.

Your turn

Task 6 - Simulating a projectile frame by frame

A game does not use the projectile equations directly. It steps the motion forward in small time steps, the same frame-by-frame update from the Motion worksheet, now with two components. In this task you write that loop in C++ and compare it with the exact answer from Task 4.

Create a new Console App (C++) project in Visual Studio named Projectile, and replace the contents of the generated source file with the skeleton below, saved as Projectile.cpp. It uses only the standard library.

1. Inside the loop, add acceleration scaled by dt to velocity, for both components.
2. Then add the new velocity scaled by dt to position, for both components. Updating velocity before position is called semi-implicit Euler.
3. After t and step are advanced, break out of the loop if position.y has dropped below zero.
4. After the loop, calculate the analytic flight time and range using the formulas from Task 4 and print them next to the simulated values. You will need to keep a copy of the launch velocity before the loop changes it.

Success condition: with dt = 0.001 the program prints a table of positions every 0.1 s, and the simulated range is within 0.05 m of the analytic 35.3 m. Then change dt to 0.1, run it again, and record how far the range is off.

#include <cmath>
#include <iomanip>
#include <iostream>

struct Vec2
{
    double x;
    double y;
};

int main()
{
    const double pi = 3.14159265358979;
    const double g = 9.81;
    const double speed = 20.0;
    const double angleDegrees = 30.0;
    const double dt = 0.001;

    const double angle = angleDegrees * pi / 180.0;
    Vec2 position{ 0.0, 0.0 };
    Vec2 velocity{ speed * std::cos(angle), speed * std::sin(angle) };
    const Vec2 acceleration{ 0.0, -g };

    const int printEvery = static_cast<int>(std::round(0.1 / dt));
    int step = 0;
    double t = 0.0;

    std::cout << std::fixed << std::setprecision(3);
    std::cout << "t (s)     x (m)     y (m)\n";

    while (t < 10.0)
    {
        if (step % printEvery == 0)
        {
            std::cout << t << "     " << position.x << "    " << position.y << "\n";
        }

        // TODO: add acceleration scaled by dt to velocity, both components

        // TODO: add velocity scaled by dt to position, both components

        t += dt;
        step++;

        // TODO: stop the loop once the projectile has dropped below y = 0
    }

    // TODO: calculate the analytic flight time T = 2 * uy / g and range
    // R = ux * T from the launch velocity, then print them next to the
    // simulated landing time t and landing distance position.x

    return 0;
}