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 of 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) Draw a position vs. time graph for both the man and bus on the same graph. Make sure to label what object each curve corresponds to and the initial position of each object in the graph.
(b) What is xman (t), the position of the man as a function of time.
(c) What is xbus (t), the position of the bus as a function of time.
(d) What condition is necessary for the man to catch the bus? Assume he catches it at the time tcatch. Hint: When are the bus and person at the same position at the same time? (e) Find the minimum speed, cmin that the man needs to catch the bus. (f) Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed c > cmin?

Answer 1

a) The position-time graph is in attachment

b) Position of the man: $x_m(t) = -b +ct$

c) Position of the bus: $x_b(t) = \frac{1}{2}at^2$

d) The condition for the man to catch the bus is $4c^2 \geq 8ab$

e) The minimum speed at which the man catches the bus is $c_{min}=\sqrt{2ab}$

f) Yes, the man can catch the bus again at $t=\frac{2c+ \sqrt{4c^2-8ab}}{2a}$

Explanation:

a)

The position-time graph for both the man and the bus are shown in the attached picture.

The coordinate x = 0 represents the initial position of the stopped bus at t = 0. Then, the bus moves with a constant positive acceleration, a, therefore its position-time graph is represented by a parabola (in a uniformly accelerated motion, the position is proportional to the square of the time).

The man starts instead at a position x = - b (distance b behind the bus). Then he continues with constant speed c, therefore its position-time graph is a straight line (constant slope means constant velocity).

b)

The motion of the man is a uniform motion with constant velocity, therefore its position at time t is given by the equation

$x_m (t) = x_0 + vt$

where

$x_0$ is the initial position

v is the velocity

For the man in this problem,

$x_0 = -b$ (initial position)

$v=c$ (velocity)

So, the position is given by

$x_m(t) = -b +ct$

c)

The motion of the bus is a uniformly accelerated motion, therefore its position at time t is given by the equation

$x_b(t)= x_0 + v_0 t + \frac{1}{2}at^2$

where

$x_0$ is the initial position

$v_0$ is the initial velocity

a is the acceleration

For this bus, we have:

$x_0 =0$

$v_0 = 0$

$a=a$

So its position is given by

$x_b(t) = \frac{1}{2}at^2$

d)

From the graph, we can see  that the man is able to catch the bus only if the slope of its line is large enough: that means, if its speed is large enough in order for him to catch the bus before the bus' velocity is too high.

In order for the man to catch the bus, the position of the man must be equal to that of the bus:

$x_m(t) = x_b(t)$

which means

$-b+ct=\frac{1}{2}at^2$

Or

$at^2-2ct+2b=0$

This is a 2nd-order equation, whose solutions are given by

$t=\frac{2c\pm \sqrt{(-2c)^2-4(2ab)}}{2a}$

Therefore, these are the two times at which the man catches the bus.

The condition for which the man can catch the bus is that the argument of the root is positive, therefore:

$(-2c)^2-4(2ab)\geq 0\\4c^2 \geq 8ab$

e)

In the previous part, we saw that the man catches the bus when

$t=\frac{2c\pm \sqrt{(-2c)^2-4(2ab)}}{2a}$

The equation has solutions if the discriminant is at least zero, therefore we can find the minimum speed by requiring the discriminant to be zero:

$\sqrt{(-2c)^2-4(2ab)} = 0\\\sqrt{4c^2-8ab}=0\\4c^2-8ab=0\\c^2=2ab\\c=\sqrt{2ab}$

So, this is the minimum speed that the man should have.

f)

By looking again at the position-time graph, we observe that the line of the man meets that of the bus twice (provided that the speed of the man is larger than the minimum speed, $c_{min}$).

These two points correspond to the two times at which the man can catch the bus, and they are the two solutions of the previous 2nd order equation:

$t_1=\frac{2c- \sqrt{4c^2-8ab}}{2a}\\t_2=\frac{2c+ \sqrt{4c^2-8ab}}{2a}$

Since $t_1$ is smaller than $t_2$, this means that if the man misses the bus at the first time of contact $t_1$, then he will be able to catch the bus again at $t_2$.

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