Q1: A
squirrel in the woods runs 2.0 m/s to the safety of a tree, trying
to escape a coyote running at 4.0 m/s. If the coyote and squirrel
start running when the coyote is 30 m away, what is the maximum
safe distance the squirrel can be from the tree to avoid becoming
lunch? Sketch and solve.
If we reflect on this problem for a moment we will realize that
both the squirrel and the coyote run for exactly the same length
of time. So time is common. Since we are given a speed (4m/s
and 2m/s) and since we need time and distance, we can use the equation:
V= d/t so, d=Vt
Time 
distance covered by coyote 
distance covered by the squirrel 
distance separating them 
1 sec 
(4m/s) x 1 sec = 4 m 
(2 m/s) x 1 sec = 2 m 
30  (42) = 28 m 
2 sec 
(4 m/s) x 2 sec = 8 m 
(2 m/s) x 2 sec = 4 m 
30  (84) = 26 m 
Notice how I am finding the distance separating the two?

30m  (distance covered by coyote  distance
covered by squirrel) 
or 
30m  (4m/s x t  2m/s x t) 

and since we are trying to find when they are
0 meters apart, we have: 

30  (4t  2t) = 0 
or 
30  4t + 2t =0 or 4t + 2t + 30
= 0 
I can rewrite my expression and get this:
4t + 2t + 30 = 0
4t + 2t = 30 and now I can solve for t,
time:
2t = 30
t= 15 seconds, meaning that at 15 seconds the squirrel and the
coyote are 0 meters apart.
So, how far can the squirrel run in 15 seconds?
V = d/t > Vt = d (2
m/s) x 15 s = d d = 30 meters
Therefore the squirrel must be less than 15 seconds (30 meters)
from a tree.
Q2: A
tourist in Yosemite runs 2.5 m/s to the safety of his car, trying
to escape a grizzly bear running at 4.0 m/s. The bear starts chasing
the tourist from 50m away. What is the maximum safe distance the
tourist can be away from his car to escape becoming an unfortunate
statistic? Sketch and solve.
Given: 
a. Tourist can run 2.5 m/s 

b. Bear can run 4.0 m/s 

c. Distance between them 50 m 


Find: 
How far can the tourist be from his car? 
This problem is identical to the first problem, so I could just
change the numbers and solve for the time.
Let's assume that the bear, man, and car are on an xaxis, with
the bear at x = 0 meters, and the man at x = 50 meters.
So the distance the man is from the bear is: (2.5 t + 50) meters.
Now I can do as I previously did.
(Distance covered by bear)  (distance covered by man) = 0 (0
is when they meet)
and using V = d/t or d = Vt I get: 4.0
m/s x t  (2.5 m/s x t + 50) = 0 or,
4t  2.5t  50 = 0
4t  2.5t = 50
1.5t = 50
t = 33.33 seconds call it 33 seconds
Since the man can run at 2.5 m/s, using d = Vt
The maximum distance the man can be is: d = (2.5)(33) = 82.5
meters from his car.
Q3: Plot
distance and velocity for situation #2.
First velocity: For both man and bear, velocity is constant.
Now for distance vs. time (Remember, velocity
is slope)
Q4: Develop
a general algebraic expression (a formula) for situations of this
sort: one object overtaking another, each with different velocities
V_{a} and V_{b}, starting a distance s apart. Your
formula needs to predict the distance d covered by the overtaken
object when it is caught. Write your formula and describe it in
two sentences.
V_{a}_{} = velocity of overtaking object, V_{b}
= velocity of overtaken object, t = time
V_{a}t  (V_{b}t + s) = 0
V_{a}t  V_{b}t  s = 0
V_{a}t  V_{b}t = s
(V_{a}  V_{b})t = s
t = s/(V_{a}  V_{b})
d of overtaken = V_{a}t = V_{b}[s/(V_{a}
 V_{b})]
The maximum distance that can be covered before being overtaken
can be found by assuming the overtaken object to be at 0 meters
and then multiplying the slower object's velocity by the ratio,
initial separation distance over the distance in velocity. Note:
The ratio must be a positive value.
