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Here is question ¿¿¿
By the method of dimensional analysis derive the relation :
S = ut + 1/2 at^2
where the letter have their usual meaning .
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Answers
Answered by
4
hey am fine dear....how are you?
Limitation of the dimensional analysis is that we can not derive the exact formula. So we can not derive S = ut +1/2at2
But we can check the correctness of the formula.
S = ut + 1/2 at2
LHS:
S = distance
unit = meter
Dimension = [M0L1T0]
RHS:
u = initial velocity
unit = m/s
Dimension = [M0L1T−1]
t = time
unit = second
Dimension = [M0L0T1]
t = time
unit = second
Dimension = [M0L0T1]
Dimension of ut = [M0L1T−1] [M0L0T1] = [M0L1T0]
This is the dimension the distance.
a = acceleration
unit = m/s2
Dimension = [M0L1T−2]
Dimension of at2 = [M0L1T−2] [M0L0T2] = [M0L1T0]
This is the dimension of distance.
Hence dimensions of the LHS and RHS are equal.
:)
Limitation of the dimensional analysis is that we can not derive the exact formula. So we can not derive S = ut +1/2at2
But we can check the correctness of the formula.
S = ut + 1/2 at2
LHS:
S = distance
unit = meter
Dimension = [M0L1T0]
RHS:
u = initial velocity
unit = m/s
Dimension = [M0L1T−1]
t = time
unit = second
Dimension = [M0L0T1]
t = time
unit = second
Dimension = [M0L0T1]
Dimension of ut = [M0L1T−1] [M0L0T1] = [M0L1T0]
This is the dimension the distance.
a = acceleration
unit = m/s2
Dimension = [M0L1T−2]
Dimension of at2 = [M0L1T−2] [M0L0T2] = [M0L1T0]
This is the dimension of distance.
Hence dimensions of the LHS and RHS are equal.
:)
Anonymous:
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Answered by
4
s = ut + 1/2 at^2
Here, s is the distance covered which is in metre,
u is the initial velocity,
t is the time taken,
a ia the acceleration.
L.H.S. is "s", which in dimensional analysis is M^0 L^1 T^0
L.H.S.= M^0 L^1 T^0
In R.H.S,
Dimension for u is L^1 T^(-1)
Dimension for t is T^1
Dimension for a is L^1 T^(-2)
Now,
R.H.S. = [M^0 L^1 T^(-1) T^1] + 1/2 [ M^0L^1 T^(-2) T^(2)]
= [M^0 L^1 T^0] + 1/2 [M^0 L^1 T^0]
R.H.S. = M^0 L^1 T^0
It is because multiplying the dimensions with 'any integer' doesn't affect it's dimensions; and addition of two dimensions which is actually the same doesn't have effect in it's dimension.
Hope this will help you.
Here, s is the distance covered which is in metre,
u is the initial velocity,
t is the time taken,
a ia the acceleration.
L.H.S. is "s", which in dimensional analysis is M^0 L^1 T^0
L.H.S.= M^0 L^1 T^0
In R.H.S,
Dimension for u is L^1 T^(-1)
Dimension for t is T^1
Dimension for a is L^1 T^(-2)
Now,
R.H.S. = [M^0 L^1 T^(-1) T^1] + 1/2 [ M^0L^1 T^(-2) T^(2)]
= [M^0 L^1 T^0] + 1/2 [M^0 L^1 T^0]
R.H.S. = M^0 L^1 T^0
It is because multiplying the dimensions with 'any integer' doesn't affect it's dimensions; and addition of two dimensions which is actually the same doesn't have effect in it's dimension.
Hope this will help you.
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