f,4 phys,,,2 balls

At t= t ,the velocity of the first ball is v
the velocity of the secendy ball is 0

(a=v/t therefore v=at
and the time equal to 3t (Let the velocity be u)
a=u/(3t) therefore u=3at )

so when the time equal to 3t ,the velocity of the first is 3v
(because u=3at=3v (at=v))

and at t=3t the velocity of the secend ball (Let the velocity is s)
=2at (at=v) (it start moveing at t=t at t=3t it moved 2t,so a=s/2t s=2at)
=2v

so the difference of the two balls at time t=3t is 3v-2v =v

做ＭＣ
假v=10ms-1，即t1=1s
求3秒後的速率變化。
波１
v=u+at
v=10+10(3)
v=40ms-1
波２
v=u+at
v=0+10(3)
v=30ms-1

波１速率－波２速率=40-30=10ms-1

由於我假設了v=10ms^-1
所以答案是v。