Question

please explain me the graphical representation of second equation of Motion and third equation of motion if possible with diagram

Answer 1

Derive s = ut + (1/2) at2 by Graphical MethodVelocity–Time graph to derive the equations of motion.Suppose the body travels a distance s in time t. In the above Figure, the distance travelled by the body is given by the area of the space between the velocity – time graph AB and the time axis OC,which is equal to the area of the figure OABC. Thus:Distance travelled=Area of figure OABC=Area of rectangle OADC + Area of triangle ABDWe will now find out the area of the rectangle OADC and the area of the triangle ABD.(i) Area of rectangle OADC=OA × OC=u × t=ut ...... (5)(ii) Area of triangle ABD=(1/2) × Area of rectangle AEBD=(1/2) × AD × BD=(1/2) × t × at (because AD = t and BD = at)=(1/2) at2...... (6)So, Distance travelled, s=Area of rectangle OADC + Area of triangle ABDor s = ut + (1/2) at2
This is the second equation of motion. It has been derived here by the graphical method.3. Derive v2 = u2 + 2as by Graphical MethodVelocity–Time graph to derive the equations of motion.We have just seen that the distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium. In other words,Distance travelled, s=Area of trapezium OABCNow, OA + CB = u + v and OC = t. Putting these values in the above relation, we get: ...... (7)We now want to eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion. Thus, v = u + at (First equation of motion)
And, at = v – u or Now, putting this value of t in equation (7) above, we get: or 2as=v2 – u2 [because (v + u) × (v – u) = v2 – u2]or v2=u2 + 2as.