Respuesta :
Answer:
(a) A = [tex][LT^{- 2}][/tex]
B = [tex][LT^{- 3}][/tex]
[tex]C = [LT^{- 5}][/tex]
(b) A = [tex]ms^{- 2}[/tex]
B = [tex]ms^{- 3}[/tex]
C = [tex]ms^{- 5}[/tex]
Solution:
The acceleration of a body is the rate at which the velocity of the body changes.
Thus
[tex]a = \frac{\Delta v_{o}}{\Delta t}[/tex]
The SI unit of velocity of an object is [tex]ms^{- 1}[/tex] and its dimension is [LT^{- 1}] and for time, T the SI unit is second, s and dimension is [T] and hence
The SI unit and dimension for the acceleration of an object is [tex]ms^{- 2}[/tex] and [LT^{- 2}] respectively.
Now, as per the question:
acceleration, a = [tex]A + Bt + Ct^{3}[/tex]
(a) Now, according to the homogeneity principle in dimension, the dimensions on both the sides of the eqn must be equal,
For the above eqn:
[tex]LT^{- 2} = A + Bt + Ct^{3}[/tex]
Thus the dimensions of :
A = [tex][LT^{- 2}][/tex]
BT = [tex][LT^{- 2}][/tex]
Thus for B
B = [tex][LT^{- 3}][/tex]
[tex]CT^{3} = LT^{- 2}[/tex]
[tex]C = [LT^{- 5}][/tex]
(b) For the units of A, B and C, we will make use of their respective dimensional formula from part (a)
where
L corresponds to length in meter(m)
T corresponds to time in seconds(s)
Now, for:
A = [tex][LT^{- 2}] = ms^{- 2}[/tex]
B = [tex][LT^{- 3}] = ms^{- 3}[/tex]
C = [tex][LT^{- 5}] = ms^{- 5}[/tex]
