User:Inconspicuum/Physics (A Level)/Graphs

There are two types of graphs of motion you need to be able to use and understand: distance-time graphs and velocity-time graphs.

Distance-time Graphs

An object travels at a constant rate for 6 seconds, stops for 5 seconds, returns to its original position in the next 7 seconds, travelling more slowly in the middle section of its return journey.

A distance-time graph plots the distance of an object away from a certain point, with time on the x-axis and distance on the y-axis.There are several types of graphs of motion you need to be able to use and understand: distance-time graphs, position-time graphs, and velocity-time graphs.

Position-time Graphs or Displacement - Time Graphs

Distance-Time Graphs give you speed, but speed is never negative so you can only have a positive slope in a distance-time graph. Position-Time graphs show displacement, have direction, and from which you can calculate velocity. If we were to imagine the line on the position-time graph to the right as a function f(t), giving an equation for s = f(t), we could differentiate this to gain:

${\frac {ds}{dt}}=f'(t)$,

where s is displacement, and t is time. By finding f'(t) at any given time t, we can find the rate at which distance is increasing (or decreasing) with respect to t. This is the gradient of the line. A positive gradient means that distance is increasing, and a negative gradient means that distance is decreasing. A gradient of 0 means that the object is stationary. The velocity of the object is the rate of change of its displacement, which is the same as the gradient of the line on a distance-time graph. This is not necessarily the same as the average velocity of the object v:

$v={\frac {s}{t}}$

Here, t and s are the total changes in displacement and time over a certain period - they do not tell you exactly what was happening at any given point in time.

Velocity-time Graphs

An object accelerates for 6 seconds, hits something, accelerates for 5 seconds and then decelerates to a stop in the remaining 6.5 seconds of its journey.

A velocity-time graph plots the velocity of an object, relative to a certain point, with time on the x-axis and velocity on the y-axis. We already know that velocity is the gradient (derivative) of the distance function. Since integration is the inverse process to differentiation, if we have a velocity-time graph and wish to know the distance travelled between two points in time, we can find the area under the graph between those two points in time. In general:

If $v=f(t)\,$

$s=\int _{t_{1}}^{t_{2}}f(t)\,dt$

where v is velocity (in ms^-1), t is time (in s), and s is the distance travelled (in m) between two points in time t₁ and t₂.

Also, by differentiation, we know that the gradient (or derivative of v = f(t)) is equal to the acceleration of the object at any given point in time (in ms^-2) since:

$a={\frac {dv}{dt}}$