Welcome to My Bitesize, let's get you set up!

Add your subjects to find the right study guides, track progress and keep everything in one place.

Add my subjects
My Subjects

Using and interpreting graphs - EduqasInstantaneous 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 ofMathsAlgebra

Instantaneous rates of change - Higher

When a relationship between two variables is defined by a curve it means that the , or 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.