Question

Consider a hot air balloon rising vertically from a launch site located on the ground. A person is initially standing 10 m from the launch site and begins walking towards the site at a rate of 3 m/s at the moment of launch. If the hot air balloon is rising at a constant rate of 4 m/s, how fast is the distance between the person and the balloon changing 2 seconds after the person starts walking?

Answer 1

The distance between the person and the balloon after 2 seconds the person starts walking is changing on 4.47 m/s

Step-by-step explanation:

The relative position between the balloon and the person is found using Galileo's relativity principle:

`\overrightarrow{r_(b/p)}=\overrightarrow{r_(b)}+\overrightarrow{r_(p)}` (1)

with Rb the position of the balloon and Rp the position of the person respect with the origin (See Figure 1). Because we don’t have those positions but we know the constant velocities, we can relate the positions (R) with the velocities (v) with the kinematic equation:

`\overrightarrow{R}=\overrightarrow{v}*t` (2)

So equation (1) is:

`\overrightarrow{R_(b/p)}=\overrightarrow{v_(b)*t}+\overrightarrow{R_(p)}=(v_(b)*t)\,\hat{j}-R_(p\,)\hat{i}` (3)

with
`\hat{j}` and
`\hat{i}` the unitary vectors on y and x direction respectively.

We see from our initial condition (See figure 2) that:

`R_(p)=(10-v_(p)*t)` (4)

So if we put this on (3) and divide by time we have:

`\frac{\overrightarrow{R_(b/p)}}{t}=\overrightarrow{v_(b/p)}=\frac{(v_(b)*\cancel{t})}{\cancel{t}}\,\hat{j}-((10-v_(p)*t))/(t)\hat{i}`(5)

But we are interested in how fast a distance is changing, and that is a speed so:

`v_(b/p)=\sqrt{v_(b)^(2)+\left(((10-(v_(p)*t)))/(t)\right)^(2)}=\sqrt{4^(2)+\left(((10-(3*2)))/(2)\right)^(2)}\simeq4.47\,(m)/(s)`