Finding distance given average velocities?

Question

Ok, I've been trying to figure out how to do this problem for an hour now and I can't take it anymore. I need some help. How the heck do I do this problem:

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 5.37 m/s. The car is a distance d away. The bear is 39.5 m behind the tourist and running at 6.54 m/s. The tourist reaches the car safely. What is the maximum possible value for d?

Answer 1

This is funny, j13562 and I got similar results using slightly different approaches. I started this a while ago and was distracted.

d/5.37 = (d + 39.5)/6.54

d/5.37 = d/6.54 + 39.5/6.54

d/5.37 - d/6.54 = 39.5/6.54

d (1/5.37 - 1/6.54) = 39.5/6.54

d  = (39.5/6.54)/(1/5.37 - 1/6.54)

d = 181.294872

d = 181.3 m should do it

(hmmm, I wonder if the bear understands rounding :-o)

You can check it out on Google by entering the following in the Google search box:

(39.5/6.54)/(1/5.37 - 1/6.54)

Answer 2

GIVEN: Vt=5.37 m/s, xb0= -39.5 m, vb=6.54 m/s, xt0=0 m

xtf=xbf

xtf and xbf both=d

vt= velocity of tourist, xb0=initial position of the bear, xt0=initial position of the tourist, vb=velocity of the bear, xtf and xbf both = d, d=the maximum distance, t^2= time squared.

Note: There is no acceleration because they are running at constant velocities

xf=x0+vt-(1/2)(a)(t^2)

xf=x0+vt-0

xf= x0+vt

xt0+vt=xb0+vb

0+5.37(t)=(-39.5 m)+(6.54 m/s)(t)

(5.37 m/s)(t)+(39.5 m)= (6.54 m/s)(t)

39.5 m=(1.17 m/s)(t)

(39.5 m) / (1.17 m/s) = t

33.8 s=t

xtf=xt0+vt

xtf=0 m+(5.37 m/s)(33.8 s)

xtf=181.5 m

Check:

xbf=x0+vb

xbf=(-39.5 m)+(6.54 m/s)(33.8 s)

xbf=181.5 m