Displacement Against Time Graphs (s/t graphs) 






In this graph: 



the lines, a, represent stationary bodies 



the lines, b, represent motion with uniform (constant) velocity 









This graph represents accelerated motion 



If the acceleration is uniform then the curve is parabolic
since
s a t^{2} 



The slope of an s/t graph represents velocity 






Velocity Against Time Graphs (v/t graphs) 






In this graph: 



the line, a, represents uniform velocity 



the lines, b, represent uniform acceleration 











The lines on this graph represent nonuniform accelerations 





The slope of a v/t graph represents acceleration 



v/t graph for a body in free fall 






Using a v/t Graph to Find Distance Moved (or
Displacement) 




Here, the body moved with constant velocity 



During the 8s period shown, it is clear that the body moved 



8 × 10 = 80m 









In this case, the body was accelerating,
uniformly, so the average speed during the 6s period was 



15/2 = 7.5ms^{1} 



therefore, the distance moved was 



7.5 × 6 = 45m 




Notice that, in each case, the arithmetic needed to find the distance moved
by the body is equivalent to finding the area between the line and the time axis. 









area of rectangle = base × height 





















area of rectangle = ½(base × height) 












In these simple cases, this observation changes nothing about the calculation
needed to find the distance moved by the
body 









However, if the shape of the graph is more complicated,
as shown here, the only way to find the distance moved might be to draw
the graph and measure the area under the curve (or more likely ask a computer to
do it for you). 







The following graph
represents a situation in which the sense of the motion of the body
changed 




Combining the areas between the graph and the
time axis, we have 




A_{1} = +12 


A_{2} = +12 


A_{3} = 1 


A_{4} = 6 



This means that the body moved
a total distance of 31m. 



However, because of the change in the sense of the
motion at t = 6s, the final
displacement of the body at t = 10s,
from its position at t = 0,
is (24  7)m = 17m in the positive sense. 
