A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 22 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?

Question

farcher · Answer

In terms of d, how long does it take the tourist to reach the car?
In terms of d, how long does it take the bear to reach the car?

If these two times are the same then the bear and the tourist reach the car at the same time.

michele_laino · Answer

the problem can be described by this drawing:
|dw:1453625321667:dw|
the eqaution of motions are:
$$\Large  \begin{gathered}
  {x_b} = {v_b}t\quad for\;bear \hfill \
   \hfill \
  {x_t} = {v_t}t + \Delta \quad for\;tourist \hfill \ 
\end{gathered} $$
where $\Delta=22$, $v_b=6$, and $v_t=3.5$
now, let's suppose that at time $\tau$ the bears reaches the tourist, then we can write:
$$\Large {v_b}\tau  = {v_t}\tau  + \Delta $$
and solving for $\tau$, we get:
$$\Large \tau  = \frac{\Delta }{{{v_b} - {v_t}}}$$
Now, if we want that the bears doesn't reach the tourist, then we have to require that the distance covered by the tourist at the time $\tau$ has to be greater or equal to $d$, namely:
$$\Large {v_t}\tau  \geqslant d$$
and after a simple substitution, we get:
$$\Large \frac{{{v_t}\Delta }}{{{v_b} - {v_t}}} \geqslant d$$