Answer :
Explanation:
The acceleration of an object is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
Dimension of velocity is, [tex][v]=[LT^{-1}][/tex]
Dimension of time is, [tex][t]=[T][/tex]
Dimension of acceleration is, [tex][a]=[LT^{-2}][/tex]
Option 1.
[tex]\dfrac{v^2}{x}=[LT^{-2}][/tex]
Option 2.
[tex]xt^2=[LT^2][/tex]
Option 3.
[tex]\dfrac{x}{t^2}=[LT^{-2}][/tex]
Option 4.
[tex]\dfrac{v}{t}=[LT^{-2}][/tex]
So, from above calculations, it is clear that option (1),(3) and (4) have the dimensions of acceleration. Hence, this is the required solution.
Answer:
a)[tex]\dfrac{v^2}{x}[/tex]
c)[tex]\dfrac{x}{t^2}[/tex]
d)[tex]\dfrac{v}{t}[/tex]
Explanation:
Acceleration :
Acceleration is the rate of change of velocity of the particle.
The unit of acceleration is m/s².
In mathematical form,
[tex]a=\dfrac{dv}{dt}[/tex]
We will check the all option on base of unit
(a). [tex]\dfrac{v^2}{x}[/tex]
Where, v = velocity
x = position
The unit of velocity and position are m/s and m.
The dimension formula of velocity and position
[tex]v = LT^{-1}[/tex]
[tex]x=L[/tex]
Put the unit in the given equations
[tex]\dfrac{v^2}{x}=\dfrac{L^2T^{-2}}{L} =\dfrac{L}{T^2}[/tex]
(b). [tex]xt^2=L\times T^2[/tex]
(c). [tex] \dfrac{x}{t^2}=\dfrac{L}{T^2}[/tex]
(d). [tex]\dfrac{v}{t}=\dfrac{LT^{-1}}{T}=\dfrac{L}{T^2}[/tex]
Hence, This is the required solution.