Derive v = u + at and s = ut + 1/2 at2 by graphical method.
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Answer:
Answer:-
Consider the velocity – time graph of a body shown in the below Figure.
Velocity – Time graph to derive the equations of motion.
Velocity-Time graph to derive the equations of motion
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 'a' from A to B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE
Now, Initial velocity of the body, u = OA ------- (1)
And, Final velocity of the body, v = BC -------- (2)
But from the graph BC = BD + DC
Therefore, v = BD + DC -------- (3)
Again DC = OA
So, v = BD + OA
Now, From equation (1), OA = u
So, v = BD + u --------- (4)
We should find out the value of BD now.
We know that the slope of a velocity – time graph is equal to acceleration, a
Thus, Acceleration, a = slope of line AB
or a = BD/AD
But AD = OC = t,
so putting t in place of AD in the above relation, we get:
a = BD/t
or BD = at
Now, putting this value of BD in equation (4) we get
v = at + u
This equation can be rearranged to give:
v = u + at
And this is the first equation of motion.
It has been derived here by the graphical method.
2. Derive s = ut + (1/2) at2 by Graphical Method
Velocity-Time graph to derive the second equation of motion
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.
3. Derive v2 = u2 + 2as by Graphical Method
Velocity-Time graph to derive the third equation of motion
Velocity–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 OABC
Distance travelled =Area of trapezium
Now, 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
This is the third equation of motion.
i hope it will helps you.......｡◕‿◕｡
V = U + at
Initial velocity u at A = OA
V changes from A to B in time t
(Uniform acceleration)
Final velocity v = BC
BC = BD +DC
V = BD + AO
V = BD + U
slope of velocity time graph is equal to acceleration
B = at
Therefore V = U + at.
S = ut + 1/2 at²
Average velocity = Total distance/time
= S/T
also ,
Average velocity = V+4/2
S/T = V+4/2.......(i)
We know that
V = U + at
Putting in 1st eq
S/T = U+ at +4 / 2
= 2U+ at/2
S/T = 4+at/2
Hence
s = ut + 1/2 at²
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hope it will help you:)
