Question

Check the correctness of the relation $s=ut+\frac{1}{2}a{t}^2,$ where $u$ is initial velocity, $a$ the acceleration, $t$ the time and $s$ the displacement.

• A. As LHS=RHS, formula is dimensionally correct.
• B. LHS and RHS doesnt have the same dimension, hence incorrect
• C. RHS is dimensionally incorrect
• D. None of these

Answer 1

The correct option is A As LHS=RHS, formula is dimensionally correct.

Writing the dimensions of either side of the given equation.
$LHS=8=displacement=\left[M^{0}L{T}^0\right]$
$RHS=ut=velocity\times time=\left[M^{0}L{T}^{-1}\right]\left[T\right]=\left[M^{0}L{T}^0\right]$
and $\frac{1}{2}a{t}^2=\left(acceleration\right)\times {\left(time\right)}^2=\left[M^{0}L{T}^{-2}\right]{\left[T\right]}^2=\left[M^{0}L{T}^0\right]$
As $LHS=RHS$, formula is dimensionally correct.