Derive second equation of motion using area of trapezium ( not area of rectangle and triangle )
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. Derive s = ut + (1/2) at2 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 ABD
We 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 ABD
or s = ut + (1/2) at2
This is the second equation of motion. It has been derived here by the 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 ABD
We 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 ABD
or 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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