Question

Check the accuracy of relation ${v^2} - {u^2} = 2as$, where v and u are the final and initial velocities, a is the acceleration, and s is the distance.

Answer 1

Hint: In this question use the concept that the accuracy is dependent upon the dimension of the relation if the L.H.S and R.H.S dimension are equal then the accuracy of the given relation is right so use this concept to reach the solution of the question.

Complete Step-by-Step solution:
So the given relation is
${v^2} - {u^2} = 2as$
Here v = final velocity and the unit of velocity is m/s.
As we know that the dimension of meter is [L] and the dimension of second is [T], so the dimension of final velocity is
$v = \dfrac{{\left[ L \right]}}{{\left[ T \right]}} = \left[ {L{T^{ - 1}}} \right]$
So the dimension of ${v^2}$ is
${v^2} = {\left[ {L{T^{ - 1}}} \right]^2} = \left[ {{L^2}{T^{ - 2}}} \right]$
Now u = initial velocity whose unit is same as final velocity as both are velocities
So the velocity of u is
$u = \dfrac{{\left[ L \right]}}{{\left[ T \right]}} = \left[ {L{T^{ - 1}}} \right]$
So the dimension of ${u^2}$ is
${u^2} = {\left[ {L{T^{ - 1}}} \right]^2} = \left[ {{L^2}{T^{ - 2}}} \right]$
Now take L.H.S of the given equation
$ \Rightarrow {v^2} - {u^2}$
Now as we see that the dimension of ${v^2}$ and ${u^2}$ are same so the dimension of $\left( {{v^2} - {u^2}} \right)$ is also same as dimension is independent of addition or subtraction
Therefore the dimension of $\left( {{v^2} - {u^2}} \right)$ is $\left[ {{L^2}{T^{ - 2}}} \right]$
So, L.H.S dimension = $\left[ {{L^2}{T^{ - 2}}} \right]$
Now consider the R.H.S of the given equation
$ \Rightarrow 2as$
Here a = acceleration.
And the unit of acceleration is m/${s^2}$.
As we know that the dimension of meter is [L] and the dimension of second is [T], so the dimension of final acceleration is
$a = \dfrac{{\left[ L \right]}}{{\left[ {{T^2}} \right]}} = \left[ {L{T^{ - 2}}} \right]$
Now s = distance and distance is measured in meter or centimeter.
And the dimension of meter or centimeter is same = [L]
So the dimension of s = [L]
So the dimension of as = $\left[ {L{T^{ - 2}}} \right]\left[ L \right] = \left[ {{L^2}{T^{ - 2}}} \right]$
Now as we know that the dimension is independent of scaling so the dimension of 2as is $\left[ {{L^2}{T^{ - 2}}} \right]$.
So R.H.S dimension = $\left[ {{L^2}{T^{ - 2}}} \right]$
So as we see that the L.H.S dimension = R.H.S dimension = $\left[ {{L^2}{T^{ - 2}}} \right]$
Hence the given relation is accurate.
So this is the required answer.

Note – That all physical quantities can be expressed in terms of seven fundamental base quantities such as mass, length, time , temperature, electric current, luminous intensity and amount of substance. These seven quantities are called seven dimensions of the physical world. We can use symbols instead of the names of the base quantities and they are represented in square brackets. [M] for mass, [L] for length, [T] for time, [K] for temperature, [I] for current, [cd] for luminous intensity and [mol] for the amount of substance. These will specify the nature of the unit and not its magnitude.