Draw velocity- time graph for a body that has initial velocity u and is moving with uniform acceleration a, Use it to derive.
a) v= u+ at
b) s= ut+ 1/2at2
c) v 2.= u2 + 2as.
Answers
Answer:
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Explanation:
Derivation of Equation of Motion
In the above graph,
The velocity of the body changes from A to B in time t at a uniform rate.
BC is the final velocity and OC is the total time t.
A perpendicular is drawn from B to OC, a parallel line is drawn from A to D, and another perpendicular is drawn from B to OE (represented by dotted lines).
Following details are obtained from the graph above:
The initial velocity of the body, u = OA
The final velocity of the body, v = BC
From the graph, we know that
BC = BD + DC
Therefore, v = BD + DC
v = BD + OA (since DC = OA)
Finally,
v = BD + u (since OA = u) (Equation 1)
Now, since the slope of a velocity-time graph is equal to acceleration a,
So,
a = slope of line AB
a = BD/AD
Since AD = AC = t, the above equation becomes:
BD = at (Equation 2)
Now, combining Equation 1 & 2, the following is obtained:
v = u + at
Derivation of First Equation of Motion by Calculus Method
Since acceleration is the rate of change of velocity, it can be mathematically written as:
a=dvdt
Rearranging the above equation, we get
adt=dv
Integrating both the sides, we get
∫t0adt=∫vudv
at=v−u
Rearranging, we get
v=u+at
Derivation of Second Equation of Motion
For the derivation of the second equation of motion, consider the same variables that were used for derivation of the first equation of motion.
Derivation of Second Equation of Motion by Algebraic Method
Velocity is defined as the rate of change of displacement. This is mathematically represented as:
Velocity=DisplacementTime
Rearranging, we get
Displacement=Velcoity×Time
If the velocity is not constant then in the above equation we can use average velocity in the place of velocity and rewrite the equation as follows:
Displacement=(InitialVelocity+FinalVelocity2)×Time
Substituting the above equations with the notations used in the derivation of the first equation of motion, we get
s=u+v2×t
From the first equation of motion, we know that v = u + at. Putting this value of v in the above equation, we get
s=u+(u+at))2×t
s=2u+at2×t
s=(2u2+at2)×t
s=(u+12at)×t
On further simplification, the equation becomes:
s=ut+12at2
Derivation of Second Equation of Motion by Graphical Method
Derivation of Equation of Motion
From the graph above, we can say that
Distance travelled (s) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD
s=(12AB×BD)+(OA×OC)
Since BD = EA, the above equation becomes
s=(12AB×EA)+(u×t)
As EA = at, the equation becomes
s=12×at×t+ut
On further simplification, the equation becomes
s=ut+12at2
Derivation of Second Equation of Motion by Calculus Method
Velocity is the rate of change of displacement.
Mathematically, this is expressed as
v=dsdt
Rearranging the equation, we get
ds=vdt
Substituting the first equation of motion in the above equation, we get
ds=(u+at)dt
ds=(u+at)dt=(udt+atdt)
On further simplification, the equation becomes:
∫s0ds=∫t0udt+∫t0atdt
=ut+12at2
Derivation of Third Equation of Motion
Derivation of Third Equation of Motion by Algebraic Method
We know that, displacement is the rate of change of position of an object. Mathematically, this can be represented as:
Displacement=(InitialVelocity+FinalVelocity2)×t
Substituting the standard notations, the above equation becomes
s=(u+v2)×t
From the first equation of motion, we know that
v=u+at
Rearranging the above formula, we get
t=v−ua
Substituting the value of t in the displacement formula, we get
s=(v+u2)(v−ua)
s=(v2−u22a)
2as=v2−u2
Rearranging, we get
v2=u2+2as
Derivation of Third Equation of Motion by Graphical Method
From the graph, we can say that
The total distance travelled, s is given by the Area of trapezium OABC.
Hence,
S = ½ (Sum of Parallel Sides) × Height
S=(OA+CB)×OC
Since, OA = u, CB = v, and OC = t
The above equation becomes
S= ½ (u+v) × t
Now, since t = (v – u)/ a
The above equation can be written as:
S= ½ ((u+v) × (v-u))/a
Rearranging the equation, we get
S= ½ (v+u) × (v-u)/a
S = (v2-u2)/2a
Third equation of motion is obtained by solving the above equation:
v2 = u2+2aS
Derivation of Third Equation of Motion by Calculus Method
We know that acceleration is the rate of change of velocity and can be represented as:
a=dvdt (1)
We also know that velocity is the rate of change of displacement and can be represented as:
v=dsdt (2)
Cross multiplying (1) and (2), we get
adsdt=vdvdt
∫s0ads=∫vuvds
as=v2−u22
v2=u2+2as
This is how we derive the three equations of motion by algebraic, graphical and calculus method.