Cambridge IGCSE Physics

Speed and velocity

Speed is the distance travelled per unit time. For a steady speed,

where $v$ is speed in $\mathrm{m\,s^{-1}}$, $s$ is distance in metres and $t$ is time in seconds. Most real trips have a speed that changes, so it is more useful to work with the average speed:

Velocity is speed in a given direction. Two cars travelling at $20\,\mathrm{m\,s^{-1}}$ in opposite directions have the same speed but opposite velocities. Speed is a scalar and velocity is a vector (the difference is explained in the note on forces and vectors).

A cyclist rides 1.2 km in 4.0 minutes and then 0.6 km in 2.0 minutes. The total distance is 1 800 m and the total time is 6.0 minutes, which is 360 s, so the average speed is $1800 \div 360 = 5.0\,\mathrm{m\,s^{-1}}$. Converting to SI units first avoids the most common slip, which is mixing kilometres with seconds.

Acceleration

Acceleration is the change in velocity per unit time.

Here $\Delta v$ is the change in velocity in $\mathrm{m\,s^{-1}}$, $\Delta t$ is the time taken in seconds and $a$ is in $\mathrm{m\,s^{-2}}$. A vehicle that speeds up from $4\,\mathrm{m\,s^{-1}}$ to $16\,\mathrm{m\,s^{-1}}$ in $3.0\,\mathrm{s}$ has $a = 12 \div 3.0 = 4.0\,\mathrm{m\,s^{-2}}$.

A deceleration is a negative acceleration: the velocity is getting smaller in the direction of travel. A train slowing from $30\,\mathrm{m\,s^{-1}}$ to $10\,\mathrm{m\,s^{-1}}$ in $8.0\,\mathrm{s}$ has $\Delta v = 10 - 30 = -20\,\mathrm{m\,s^{-1}}$, so $a = -20 \div 8.0 = -2.5\,\mathrm{m\,s^{-2}}$. Keep the minus sign when it comes out of the calculation, because it tells you the object is slowing.

Distance-time graphs

On a distance-time graph the vertical axis is distance and the horizontal axis is time, so the gradient is distance divided by time, which is speed.

To find the speed from a straight section, take two well-separated points on it and work out the rise divided by the run. A steeper line means a greater speed. A distance-time graph can never slope downwards when distance is the total length travelled, because distance travelled does not decrease.

Speed-time graphs

On a speed-time graph the vertical axis is speed, so the gradient is change in speed divided by time, which is acceleration.

The area under a speed-time graph is the distance travelled. For constant speed the area is a rectangle, and for constant acceleration from rest or between two speeds it is a triangle, a trapezium or a combination of these.

A worked graph

An object accelerates uniformly from rest to $12\,\mathrm{m\,s^{-1}}$ in 6 s, travels at $12\,\mathrm{m\,s^{-1}}$ for 10 s, then decelerates uniformly to rest in 4 s.

The acceleration in the first stage is the gradient, $12 \div 6 = 2.0\,\mathrm{m\,s^{-2}}$. The distance in each stage is the area:

The total distance is $36 + 120 + 24 = 180\,\mathrm{m}$ and the total time is 20 s, so the average speed is $180 \div 20 = 9.0\,\mathrm{m\,s^{-1}}$. The gradient in the last stage is $-12 \div 4 = -3.0\,\mathrm{m\,s^{-2}}$.

When the line is a curve the acceleration is changing. Extended candidates should be able to say whether the acceleration is constant (a straight line) or changing (a curve) and, for a curve, whether it is increasing or decreasing from how the gradient changes.

Free fall

Near the surface of the Earth every object in free fall has the same acceleration, called the acceleration of free fall, $g$, which is approximately $9.8\,\mathrm{m\,s^{-2}}$ and approximately constant over the heights we deal with. A dense object and a light one dropped together hit the ground together when air resistance is negligible, because the extra weight of the heavy one is cancelled by its extra mass, which resists acceleration in proportion.

A ball dropped from rest has a speed of about $9.8\,\mathrm{m\,s^{-1}}$ after 1 s, $19.6\,\mathrm{m\,s^{-1}}$ after 2 s and so on, so its speed-time graph is a straight line through the origin with gradient $g$.

Falling with air resistance and terminal velocity

Extended only: an object falling through air or a liquid feels a drag force upwards that grows as its speed increases. The motion of a skydiver, or of a ball bearing dropped into a tall cylinder of oil, has three stages.

1. At the start the speed is small, so the drag is small and the weight is much bigger. The resultant force is nearly the whole weight and the acceleration is close to $g$.
2. As the speed rises the drag rises too, so the resultant force (weight minus drag) gets smaller and the acceleration decreases. The speed-time graph curves and becomes shallower.
3. Eventually the drag equals the weight. The resultant force is zero, so the acceleration is zero and the object falls at a constant speed, called its terminal velocity. The graph has become horizontal.

If a skydiver opens a parachute, the area facing the air suddenly increases, so the drag becomes larger than the weight. The resultant force is now upwards and the skydiver decelerates, which lowers the speed and so the drag, until the two forces balance again at a new and much smaller terminal velocity.