# Question 32e56

Oct 22, 2016

A particle having initial velocity u , moving with uniform acceleration a covers a distance ${S}_{t}$ in t th sec.

Let us recall the deduction of the relation among these quantities considering the average velocity of the particle for time $t \in \left[\left(t - 1\right) , t\right]$

Velocity of the particle after (t-1) sec$= {v}_{\text{t-1}} = u + a \left(t - 1\right)$

Velocity of the particle after t sec$= {v}_{t} = u + a t$

Average velocity for $t \in \left[\left(t - 1\right) , t\right]$ is v_"avg"=(v_("t-1")+v_t)/2

$\implies {v}_{\text{avg}} = \frac{2 u + a \left(2 t - 1\right)}{2}$
$\implies {v}_{\text{avg}} = u + \frac{1}{2} a \left(2 t - 1\right)$

So ${S}_{t} = {v}_{\text{avg"xx t^" th" sec=v_"avg}} \times 1 \sec$

${S}_{t} = u + \frac{1}{2} a \left(2 t - 1\right)$

Apparently the equation appears false according to dimension method. Since its LHS has dimension $\left[L\right]$ and RHS has dimension $\left[L {T}^{-} 1\right]$ .
Here ${S}_{t}$ represents distance traversed by the particle in t th sec which is actually a time interval of 1 sec. This 1 sec is multiplied in RHS and not visible as variable(t). The dimension of RHS is $\left[L {T}^{-} 1\right]$without this invisible t.

If it is multiplied by dimension of time $\left[T\right]$ for 1 sec, then the satisfaction of dimension will be found.