Derive the second equation of motion by graphical method .
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Second Equation Of Motion
Derive s = ut + (1/2) at 2 by Graphical Method Velocity–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.
Derive s = ut + (1/2) at 2 by Graphical Method Velocity–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.
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justinMars33:
can you please draw the graph
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Hope that u understood
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