Question

Bug P starts at 0 and moves along a number line in the positive
direction. Bug N starts at 0 and moves in the negative direction. Each
bug moves at a constant speed, but one bug travels faster than the other.
Bugs P and N leave 0 at the same time, and four seconds later P is at 12
and N is at –8

a) Suppose Bug P reaches point x in t seconds, and Bug N reaches
point y in t seconds. Express x in terms of t. Then express y in terms
of t.
b) How far apart are bugs P and N thirty seconds after leaving point 0?
c) When will bugs P and N be 300 units apart?
d) Draw a graph that shows how the distance between the bugs depends on
the amount of time that has passed.
e) Is the distance between the bugs directly proportional to time, inversely
proportional to time, or neither?

Answer 1

By getting the motion equations for the bugs, we will see that:

a)

• x = (3 units per seg)*t
• y = (-2 units per seg)*t

b) 150 units.

c) 60 seconds

d) The graph can be seen at the end.

e) Directly proportional.

Getting the motion equations:

a) We know that bug P moves 12 units in 4 seconds, then its velocity is:

Vp = 12/4s = 3 units per seg.

And bug N moves -8 units in 8 seconds, then its velocity is:

Vn = -8/4s = -2 units per seg.

Then we have:

• x = (3 units per seg)*t
• y = (-2 units per seg)*t

b) The distance between the bugs at the time t is:

d = x - y

At 30 segs, that distance is:

d(30s) = (3 units per seg)*30s - (-2 units per seg)*30s = 150 units.

c) Now we need to write:

d = 300 units =  (3 units per seg)*t - (-2 units per seg)*t

300 units = (5 units per seg)*t

(300 units)/(5 units per seg) = t = 60s

The distance between the bugs will be 300 units after 60 seconds.

d) The graph of:

d(t) = (5 units per seg)*t

Can be seen at the end.

e) As the time increases also does the distance, so these two are proportional.

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