# Uniformly accelerated motion represented graphically

Uniformly accelerated motion represented graphically

Fig. 3.2 shows the velocity-time graph for a body which starts with a velocity of 3 m/s and moves with an acceleration of 2 m/s 2 for 4 s. In this case the velocity increases uniformly, and therefore the average velocity is

equal to the velocity at half time and is represented by M D on the graph. If we take the right-angled triangle MBF and place it in the position MAE we shall obtain a r tangle OEFC whose area is equal to that of OABC. The height of rectangle OEFC represents the average velocity and its base repre ents the time. Therefore,

hence also
area of OEFC = average velocity x time = distance
area of 0ABC = distance (numerically)
An alternative way of looking at this problem is to take the average velocity as being equal to half the sum of the initial and final velocities, i.e., i{OA + CB) = ~ 3 + 11) = 7 m/s.
Hence, distance moved = i{OA + CB) x OC = 7 x 4 = 28 m But this expression is also equal to the area of OABC, since OABC is a trapezium and the area of any trapezium = 1{sum of parallel sides) x perpendicular distance between them. So, whichever way we look at the problem, the distance moved is seen to be numerically equal to the area between the velocity-time curve and the time axis.