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Using and interpreting graphs - EdexcelInstantaneous rates of change - Higher

Using graphs is not just about reading off values. In real-life contexts, the intercept, gradient and area underneath the graph can have important meanings such as a fixed charge, speed or distance.

Part of MathsAlgebra

Instantaneous rates of change - Higher

When a relationship between two variables is defined by a curve it means that the gradient, or rate of change is always varying.

An average speed for a journey can be found from a distance-time graph by working out the gradient of the line between the two points of interest. This is like using a car's milometer and clock to take readings at the start and end of a journey.

However, they do not tell you the actual speed at a particular moment. You need a speedometer for that. On a graph, this can be estimated by drawing a to the curve and calculating its gradient.

Example

This distance-time graph shows the first ten seconds of motion for a car.

Distance vs time graph, showing change  in rate of speed

The average speed over the 10 seconds = gradient of the line from (0, 0) to (10, 200) = \( \frac{200}{10} = 20 \:\text{m/s}\).

To find an estimate of the speed after 6.5 seconds, draw the tangent to the curve at 6.5.

\(\text{Gradient} = \frac{140 - 20}{9 - 4} = \frac{120}{5} = 24 \:\text{m/s}\)

A velocity-time graph shows the velocity of a moving object on the vertical axis and time on the horizontal axis.

The gradient of a velocity time graph represents acceleration, which is the rate of change of velocity. If the velocity-time graph is curved, the acceleration can be found by calculating the gradient of a tangent to the curve.

Example

The velocity of a sledge as it slides down a hill is shown in the graph.

Find the acceleration of the sledge when t = 6s.

Diagram to find the acceleration of the graph when t = 6s

Draw a tangent to the curve at the point where t = 6s and draw two lines to form a right angle triangle. The acceleration is equal to the gradient of the tangent which is \(\frac{change~in~y}{change~in~x} = \frac{7~m/s}{8~s} = 0.875~{m/}{s}^2\)

Notice, that after about 10 seconds, the gradients are negative meaning the sledge is slowing down or decelerating.