i found out that:
x(man)t(catch)=x(bus)t(catch)
and
-b+ct(catch)=(1/2)at^2(catch)
A man is running at speed c (much less than the speed of light) to catch a bus already at a stop. At t=0 , when he is a distance b from the door to the bus, the bus starts moving with the positive acceleration a .
Use a coordinate system with x=0 at the door of the stopped bus.
So,
What is the minimum value of c for which the man will catch the bus?
Express the minimum value for the man's speed in terms of a and b ?
PLEASE clear explanations...thanks
What is the minimum value of c for which the man will catch the bus?
Let's start with the man having constant velocity
x(t)=-b+c*t
the bus has
x(t)=.5*a*t^2
when the x(t)'s are equal, we have a catch
-b+c*t=.5*a*t^2
this is a quadratic, if there is a catch, there will be a real, positive root
0=.5*a*t^2-c*t+b
the roots are
(c^2+/-sqrt(c^2-2*a*b))/a
the smaller root will be
(c^2-sqrt(c^2-2*a*b))/a
The larger root occurs if the man runs past the bus and the bus accelerates back to him.
This requires that
(c^2-2*a*b)%26gt;=0
j
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