Question

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.

a)

What is xman(t), the position of the man as a function of time?

Answer symbolically in terms of the variables b, c, and t.

xman(t) =

b)

What is xbus(t), the position of the bus as a function of time?

Answer symbolically in terms of a and t.

xbus =

c)

What condition is necessary for the man to catch the bus? Assume he catches it at time tcatch.

What condition is necessary for the man to catch the bus? Assume he catches it at time .

	xman(tcatch)>xbus(tcatch)
	xman(tcatch)=xbus(tcatch)
	xman(tcatch)<xbus(tcatch)
	c=a?tcatch

Answer 1

SOLUTION :

Using a coordinate system with x=0 at the door of the stopped bus.

Initial position of man at t=0 is xo=-b

speed of man = v = c

Motion of man is uniform (without acceleration) with constant velocity 'c' in positive direction

A.

The position of the man as a function of time is given by x = xo + vt

x_man_(t)= - b+ c t

B.
Bus starts from origin,
Initial position of the bus is: xo=zero

Bus starts from rest, initial velocity of bus = vo = 0

Acceleration of bus= a =a

For motion with constant acceleration, the equation of motion is : x = xo+vot +(1/2)at^2

The motion of bus is with constant acceleration, the equation of motion of bus is :

x_bus_(t) = zero +zero +(1/2)at^2

x_bus_(t) = (1/2)at^2
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C . Assume he catches it at time t_catch. The possible answers for the condition necessary for the man to catch the bus are

x_man_(t_catch_) = x_bus_(t_catc...