A particle having initial velocity u , moving with uniform acceleration a covers a distance S_tSt in t th sec.
Let us recall the deduction of the relation among these quantities considering the average velocity of the particle for time tin[(t-1),t]t∈[(t−1),t]
Velocity of the particle after (t-1) sec=v_("t-1")=u+a(t-1)=vt-1=u+a(t−1)
Velocity of the particle after t sec=v_t=u+at=vt=u+at
Average velocity for tin[(t-1),t]t∈[(t−1),t] is v_"avg"=(v_("t-1")+v_t)/2vavg=vt-1+vt2
=>v_"avg"=(2u+a(2t-1))/2⇒vavg=2u+a(2t−1)2
=>v_"avg"=u+1/2a(2t-1) ⇒vavg=u+12a(2t−1)
So S_t=v_"avg"xx t^" th" sec=v_"avg"xx1secSt=vavg×t thsec=vavg×1sec#
S_t =u+1/2a(2t-1) St=u+12a(2t−1)
Apparently the equation appears false according to dimension method. Since its LHS has dimension [L][L] and RHS has dimension [LT^-1] [LT−1] .
Here S_t St 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 [LT^-1] [LT−1]without this invisible t.
If it is multiplied by dimension of time [T][T] for 1 sec, then the satisfaction of dimension will be found.
