Question
Q A PARTICLE STARTS MOVING WITH ACC.=2m/s 2 DISTANCE TRAVELLED BY IT IN 5th HALF SECOND IS=?
Q A PARTICLE MOVES IN A STRAIGHT LINE & ITS POSITION X AT TIME t IS GIVEN BY X RAISED TO P. 2=2+t . ACC. IS GIVEN BY=? (ans . -2/4t cube)
first calculate distance of 4 seconds by using s = ut + 1/2 a t2
then distance of 4.5 sec. by same formula
difference of these 2 is the distance of 5th half sec.
but some writter consider 5th half as 1/2 +1/2+1/2+1/2+1/2 means 2.5
so they use the same formula first for 2 sec then for 2.5 sec, then their difference.
according to my opinion first is more true way to solve.
Dear Sarika please submit only one question in a box, for a new question use another box so that i could give answers more quickly.
According to the given condition
x2 = 2+t
on differentiaing we get 2x dx/dt = 1
or v = 1/2x
now differentiate it again we get dv/dt = -(1/2x2 )(dx/dt) = -(1/2x2 )(-1/2x) = 1/4x3