Prove the time independent equation of motion graphically.
Answers
Answer:
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Explanation:
Graphical method : Consider an object moving along a straight line with uniform acceleration a let u be the initial velocity of the object at time t=0 and v be the final velocity of the object at time t . let s be the distance travelled by the object in time t.
velocity -time graph of this motion is a straight line PQ as shown in the figure
where OP=u=RS
OW=SQ=v and
OS=PR=t
(I) for the first equation of motion : we know that the slope of velocity -time graph of uniformly accelerated motion represents the acceleration of the object
i.e. acceleration = slope of the velocity -time graph PQ
or a=
PR
QR
=
OS
QR
a=
OS
SQ−SR
=
t
v−u
or v−u=at
or v=u+at .....(i)
This is the first equation of uniform acceterated motion.
(ii) Second equation of motion ; We know that the area under the velocity-time graph for a given time interval represents the distance covered by the uniformly accelerated object in that interval of time.
∴ Distance (displacement) travelled by the object in time t is :
S = area of trapezium OSQP
= area of rectangle OSRP + Area of triangle PRQ
or S=OS×OP+
2
1
×PR×PQ
(area of rectangle = length × breadth)
( Area of triangle =
2
1
× Base × Height )
=t×u+
2
1
×t×(v−u)
(from the first equation of motion v−u=at)
=ut+
2
1
×t×at
Thus S=ut+
2
1
at
2
......(ii)
This is the second equation of uniform accelerated motion.
(iii) Third equation of motion : Distance travelled by the object in time interval t is s = area of trapezium OSQP
=
2
1
(OP+SQ)×OS
∵OP=SR
=
2
1
(SR+SQ)×OS........(iii)
Acceleration , a = slope of the velocity -time graph PQ
or a=
PR
RQ
=
OS
SQ−SR
or OS=
a
SQ−SR
...........(iv)
Putting this value in equation (iii) we get
s=
2
1
(SR+SQ)(
a
SQ−SR
)
or s=
2a
1
(AQ
2
−SR
2
)
or s=
2a
1
(v
2
−u
2
)
or v
2
−u
2
=2as
or v
2
=u
2
+2as.............(v)
This the third equation of uniform accelerated motion.
