Question #32e56

1 Answer
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 tin[(t-1),t]

Velocity of the particle after (t-1) sec=v_("t-1")=u+a(t-1)

Velocity of the particle after t sec=v_t=u+at

Average velocity for tin[(t-1),t] is v_"avg"=(v_("t-1")+v_t)/2

=>v_"avg"=(2u+a(2t-1))/2
=>v_"avg"=u+1/2a(2t-1)

So S_t=v_"avg"xx t^" th" sec=v_"avg"xx1sec#

S_t =u+1/2a(2t-1)

Apparently the equation appears false according to dimension method. Since its LHS has dimension [L] and RHS has dimension [LT^-1] .
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 [LT^-1] without this invisible t.

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