Explanation:
(a) Velocity is given by :
[tex]v=\dfrac{ds}{dt}[/tex]
s is the length of the distance
t is the time
The dimension of v will be, [tex][v]=[LT^-1][/tex] Â Â Â
(b) The acceleration is given by :
[tex]a=\dfrac{dv}{dt}[/tex]
v is the velocity
t is the time
The dimension of a will be, [tex][a]=[LT^{-2}][/tex]
(c) Since, [tex]d=\int\limits{v{\cdot}dt} =[LT^{-1}][T]=[L][/tex]
(d) Since, [tex]v=\int\limits{a{\cdot}dt} =[LT^{-2}][T]=[LT^{-1}][/tex]
(e)
[tex]\dfrac{da}{dt}=\dfrac{[LT^{-2}]}{[T]}[/tex]
[tex]\dfrac{da}{dt}=[LT^{-3}]}[/tex]
Hence, this is the required solution.
