Velocity-Time Graphs
In the previous section, the focus was on calculations covering Speed and Acceleration. In this section, graphs of Velocity against Time will be used find information about an object's motion. An example of a Velocity-Time graph is shown below:-
In order to fully explain the use of a Velocity-Time graph, each section above will be looked at separately.
Start to A
In this first section of the graph, The object starts at rest and then increases in Velocity to 20 ms-1. By using the graph to gather data, it is possible to calculate the Acceleration at this point:-
a = ?
v = 20 ms-1
u = 0 ms-1
t = 5 s
a = (20 - 0)/5
a = 4 ms-2
Note - On a Velocity-Time graph, a diagonal line with a positive gradient shows a positive acceleration.
A to B
In this section of the graph, the object travels at a constant Velocity of 20 ms-1.
Note - On a Velocity-Time graph, a horizontal line shows a constant Velocity.
B to C
In this section, the object starts at 20 ms-1 and then experiences a negative Acceleration. The object's Velocity decreases until it reaches a Velocity of -7 ms-1. By using the graph to gather data, it is possible to calculate the negative Acceleration at this point:-
a = ?
v = -7 ms-1
u = 20 ms-1
t = 5 s
a = (-7 - 20)/5
a = -5.4 ms-2
Note - On a Velocity-Time graph, a diagonal line with a negative gradient shows a negative Acceleration.
Below X-Axis Motion
In the section B to C, the Velocity reached zero and then became negative. From an understanding of Speed, this makes no sense, but from an understanding of Velocity this is easily explained.
In this question, it was assumed that the object was initially moving in a forward direction. So when the Velocity became negative, this simply means that the object is now moving backwards.
Note - In Physics (by convention) the following applies:-
Positive direction (+) : Upwards or to the Right
Negative direction (-) : Downwards or to the Left
Distance / Displacement Traveled
By using the graph above it is also possible to calculate the Distance or Displacement an object has travelled. As these are Velocity-Time graphs, if the area under the graph is found, this will give the distance travelled.
Note - This is due to distance being equal to V x T, so the area gives distance.
The following diagram shows the graph above, colour-coded for ease of calculation:-
Note - To find area :-
Rectangle - Base x Height
Triangle - 1/2 x Base x Height
By finding the area of each section and adding them all together, the total distance can be found :-
Total Distance = Area 1 + Area 2 + Area 3 + Area 4
Total Distance = (1/2x5x20) + (10x20) + (1/2x3x20) + (1/2x2x7)
Total Distance = 287 m
To find the Displacement, add together all sections above the axis, and subtract all sections below the axis:-
Displacement = ( Area 1 + Area 2 + Area 3 ) - ( Area 4 )
Displacement = [ (1/2x5x20) + (10x20) + (1/2x3x20) ] - (1/2x2x7)
Displacement = 273 m in the forward direction
Note - Displacement is a Vector, therefore must include a direction.
