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

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Solution

We have $v^{2} - u^{2} = 2 a s$ .

Checking the dimensions on both sides, we get
$L H S = [L T^{- 1}]^{2} - [L T^{- 1}]^{2}$

$= [L^{2} T^{- 2}] - [L^{2} T^{- 2}] = [L^{2} T^{- 2}]$

$R H S = [L^{1} T^{- 2}] [L] = [L^{2} T^{2}]$

Comparing LHS and RHS, we find LHS = RHS.
Hence, the formula is dimensionally correct.

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Check the accuracy of relation v2−u2=2as, where v and u are final and initial velocities, a is the acceleration, and s is the distance.

We have v2−u2=2as.Checking the dimensions on both sides, we getLHS=[LT−1]2−[LT−1]2=[L2T−2]−[L2T−2]=[L2T−2]RHS=[L1T−2][L]=[L2T2]Comparing LHS and RHS, we find LHS = RHS.Hence, the formula is dimensionally correct.

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