What is the minimum value of c for which the man will catch the bus?

Question

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

Answer 1

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)>=0

j

Answer 2

What is the actual answer?