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.

What is x_{\rm man}(t), the position of the man as a function of time?
Answer symbolically in terms of the variables b, c, and t.

-b+c*t

To determine the position of the man as a function of time, we can consider the motion of the man and the bus separately.

For the man:

Since the man is running at a constant speed c, his position as a function of time can be expressed as:

x_man(t) = b + ct

Here, b represents the initial distance of the man from the door of the bus.

For the bus:

The bus starts moving with positive acceleration a. We can use the equation of motion for uniformly accelerated motion to describe the position of the bus as a function of time. However, since the bus starts from rest, we also need to consider the initial velocity of the bus.

Using the equation of motion:

x_bus(t) = x_0 + v_0*t + (1/2)*a*t^2

Where x_0 is the initial position of the bus (0 in this case), v_0 is the initial velocity of the bus (0 in this case), and a is the acceleration of the bus.

Combining the equations for the man and the bus:

x_man(t) = b + ct
x_bus(t) = (1/2)*a*t^2

Since the man starts at a distance b from the door, his position relative to the bus is:

x_relative(t) = x_man(t) - x_bus(t)
x_relative(t) = b + ct - (1/2)*a*t^2

Thus, the position of the man as a function of time is:

x_man(t) = b + ct - (1/2)*a*t^2

To determine the position of the man as a function of time, we need to break the problem down into two parts: the time when the bus is at rest and the time when the bus is in motion.

1. When the bus is at rest (t < t1):
During this time, the man is running towards the bus at a constant speed c. The position of the man can be calculated using the formula:

x_{\rm man}(t) = b - c * t

Here, b represents the initial distance of the man from the door of the bus, and c is the constant speed at which the man is running.

2. When the bus is in motion (t >= t1):
When the bus starts moving with a positive acceleration a, we need to find the time t1 when the man reaches the bus. To find t1, we can use the formula for constant acceleration:

d = v_0 * t + (1/2) * a * (t^2)

Where v_0 is the initial velocity of the bus (which is zero in this case), d is the distance traveled by the bus, and t is the time taken.

Since the bus is initially at rest and starts moving in the positive x direction, the distance traveled by the bus is equal to b. Plugging these values into the equation and solving for t, we get:

b = (1/2) * a * (t1^2)

Solving for t1:

t1 = sqrt((2 * b) / a)

Once we have t1, we can calculate the position of the man as a function of time when the bus is in motion:

x_{\rm man}(t) = b + c * (t - t1)

Here, t - t1 represents the time elapsed after the man reaches the bus.

So, the position of the man as a function of time, symbolically, is:

x_{\rm man}(t) = {
b - c * t, for t < t1,
b + c * (t - t1), for t >= t1
}

f(t)=ct