A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 m/s. The car is a distance away. The bear is 26m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible values for d?

Question

theeric · Answer

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 m/s. $v_\text{tourist to car}=4.0\ [m/s]$

The car is a distance away. $d_\text{tourist to car}=?$

The bear is 26m behind the tourist $d_\text{bear to tourist}=26\ [m]$
and running at 6.0 m/s. $v_\text{bear to car}=6.0\ [m/s]$

Bear is running at $6.0\ [m/s]$ relative to the ground and car. We want to see how many meters the bear gets closer to the tourist each second. Relative to the tourist, the bear runs the difference between its velocity and the tourist's, so $v_\text{bear to tourist}=v_\text{bear to car}-v_\text{tourist to car}\=6.0\ [m/s]-4.0\ [m/s]\=2.0\ [m/s]$   Getting 2 meters close to the tourist per second. Do you agree?

The tourist reaches the car safely. $\text{Tourist reaches car in less time}\\text{ than bear reaches tourist.}$
So 
$t_\text{tourist to car}<t_\text{bear to tourist}$

$v=\dfrac{d}{t}\implies t=\dfrac{d}{v}$

Then $t_\text{tourist to car}<t_\text{bear to tourist}$
$\implies \dfrac{d_\text{tourist to car}}{v_\text{tourist to car}}\lt \dfrac{d_\text{bear to tourist}}{v_\text{bear to tourist}}\~\\implies d_\text{tourist to car}\lt \dfrac{d_\text{bear to tourist}~v_\text{tourist}}{v_\text{bear to tourist}}\~\\implies d_\text{tourist to car}\lt \dfrac{26\ [m]~4.0\ [m/s]}{2.0\ [m/s]}=\dots\ [m]$

What does that mean?

Do you have any questions?