Show by using the graphical method that: s = u t + 1 2 a t 2 where the symbols have their usual meanings.

Consider the velocity-time graph of a body shown in the figure. The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t . In other words, there is a uniform acceleration from A to B , and after time t its final velocity becomes v which is equal to B C in the graph. To complete the figure, we draw the perpendicular C B from point C , and draw A D parallel to O C . B E is perpendicular from point B to O E . Suppose the body travels a distance s in time t . In figure, the distance travelled by the body is given by the area of the space between the velocity graph A B and the time axis O C , which is equal to the area of figure O A B C . Thus: s = Area of figure O A B C =Area of rectangle O A D C + Area of triangle A B D We will now find out the area of the rectangle O A D C and area of the triangle A B D Area of rectangle O A D C = O A × O C ( s e e f i g u r e ) = u × t = u t Area of triangle A B D = 1 2 × A r e a o f r e c tan g l e A E B D = 1 2 × A D × B D = 1 2 × t × a t ( b e c a u s e A D = t a n d B D = a t ) = 1 2 a t 2 So, distance travelled, s = Area of rectangle O A D C + Area of triangle A B D s = u t + 1 2 a t 2 This is the second equation of motion. It has been derived from the graphical method.