Interpreting Graphs and Kinematics in Context

Graphs in real-world contexts allow us to find approximate solutions to complex problems by visualising the relationship between variables. Understanding the specific characteristics of different graph types is essential for interpreting these contexts accurately.

Straight Line Graphs

Straight line graphs that pass through the origin ($y = mx$) represent a simple proportional relationship. For example, the cost of items when there is no bulk discount or the distance travelled at a constant speed. If the line does not pass through the origin ($y = mx + c$), the intercept $c$ often represents an initial or fixed charge, followed by a proportional relationship represented by the gradient $m$.

Example: Poster Printing

The cost of printing a poster consists of an initial set-up fee plus a fixed cost for each poster printed. We can use the graph below to find these values.

To find the set-up cost, we look at the cost when 0 copies are printed. From the graph, this intercept is $\pounds 3$. To calculate the unit cost (the cost per poster), we find a point the graph passes through clearly, such as $(10, 9)$. The cost for these 10 copies, excluding the set-up fee, is $\pounds 9 - \pounds 3 = \pounds 6$. Therefore, the unit cost is $\pounds 6 \div 10 = \pounds 0.60$ per poster. Using this, the cost for 500 posters would be the set-up fee plus the unit cost: $\pounds 3 + (500 \times \pounds 0.60) = \pounds 303$.

Reciprocal Graphs

Reciprocal graphs of the form $y = \frac{k}{x}$ are used to model inverse proportion, where one variable increases as the other decreases such that their product remains constant ($xy = k$).

Example: Gas Pressure and Volume

The graph below shows the pressure in atmospheres, $P$, plotted against the volume, $V$, of a gas in litres.

To deduce if $V$ is inversely proportional to $P$, we check if $V = \frac{k}{P}$, or $PV = k$. From the graph, when $P = 1.0$, $V = 2.5$, suggesting $k = 2.5$. We verify this for other points:

1. At $P = 2.5, V = 1.0$, so $PV = 2.5$.
2. At $P = 5.0, V = 0.5$, so $PV = 2.5$.
3. At $P = 6.0, V \approx 0.42$, so $PV \approx 2.5$.

Since $PV$ is constant, we can conclude $V = \frac{2.5}{P}$.

Exponential Graphs

Exponential curves of the form $y = k^x$ model growth or decay where a quantity changes by a constant factor over equal time periods. Common examples include population growth or radioactive decay.

Example: Bacteria Growth

The relationship between the number of bacteria $n$ (in thousands) and time $t$ (in hours) is $n = k^t$.

At the start ($t = 0$), $n = 1$, which represents 1000 bacteria. After 1 hour, $n = 3$; after 2 hours, $n = 9$. We can see the population is multiplied by 3 every hour, so $k = 3$. The equation is $n = 3^t$. To find the population at the end of 4 hours, we calculate $3^4 = 81$, which is 81,000 bacteria.

Kinematics: Distance Time Graphs

Non-standard graphs often represent kinematic problems. In a distance-time graph, the vertical axis represents distance from a fixed point and the horizontal axis represents time.

1. A straight line indicates a constant speed. The gradient of the line is the speed.
2. A horizontal line indicates the object is stationary.
3. A curve indicates non-constant speed (acceleration or deceleration).

Example: Ania's Journey

The graph shows Ania's journey from home to school, 1 km away.

Ania stops at a shop to wait for a friend. The graph is flat between 6 and 7 minutes, so she waits for 1 minute. The shop is 0.5 km (500 m) from home. Her walking speed to the shop is $500 \text{ m} \div 6 \text{ min} = 83\frac{1}{3} \text{ m/min}$. The journey from the shop to school is curved, meaning she is not moving at a constant speed; the increasing gradient shows she is accelerating. Her average speed for this 500 m leg over 3 minutes is $500 \div 3 = 166\frac{2}{3} \text{ m/min}$.