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鈥檚 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 A straight line that just touches a point on a curve. A tangent to a circle is perpendicular to the radius which meets the tangent. to the curve and calculating its gradient.
Example
This distance-time graph shows the first ten seconds of motion for a car.
The average speed over the 10 seconds = gradient of the line from (0, 0) to (10, 200) = \(\frac{200}{10} = 20~m/s\).
To find an estimate of the speed after 6.5 seconds, draw the tangent to the curve at 6.5.
\(gradient = \frac{140-20}{9-4} = \frac{120}{5} = 24~m/s\)
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.
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 \(\frac{change~in~y}{change~in~x} = \frac{7 m/s}{8s} = 0.875 m/s^2\).
Notice, that after about 10 seconds, the gradients are negative meaning the sledge is slowing down or decelerating.