oh shit
k here's d relation
at^2+bt+c=x
A particle is moving along x-axis as ; where x is in m, t in second, a = 1; b =- 3, c = 2, distance cover by particle from t = 1 to t = 2,
(A) 0.5 m (B) zero
(C) 2 m (D) None
Mata ji yahan par
A particle is moving along x-axis as_______________________ ; where x is in m, t in second, a = 1; b =- 3, c = 2, distance cover by particle from t = 1 to t = 2,
(A) 0.5 m (B) zero
(C) 2 m (D) None
Fill in the blank to kar do
dunno post d sol ............... really itz bettr then zer o :D
method:
find where velcity becomes zero
at t=3/2s
find x at t=3/2
and find 2x
can u xplain in wrds .i mean wats d logic behind this
sorry if m being dumb
initially at t=1 x=0
at t=2 also x=0
so it has returned to its initial position. Now if it returns velocity has to become zero at some point before it changes direction. so find v=dx/dt and put v=0
I know ive explained clumsily. Say if you can get or ill post again