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1 vote
On the number line, a slow bug S and a fast bug F are moving in the positive direction from the point 0. The speeds of both bugs are constant. Both leave 0 at the same time, and four seconds later F has reached 12 and S has reached 8.

If x is the point reached by F after t seconds, and y is the point reached by S after t seconds, express x and y in terms of t. x=, y=

30 seconds after leaving point 0, how far apart are F and S ? ---------- units

When will F and S be 100 units apart? ------------ seconds

Please help!!!!!!! Please don't write a bunch of nonsense for the points!!!!!!!!! Please answer all the questions!!!!!!!!!!!!!! I beg!!!!!!!!!!!!!!!!1

2 Answers

4 votes

Answer:

100 units for 100 seconds.

30 units for 30 seconds.

Explanation:

Look up

User Chandan H
by
8.4k points
3 votes
a) The speed of F is 12 units in 4 seconds, or 3 units per second.
x = 3t

The speed of S is 8 units in 4 seconds, or 2 units per second.
y = 2t

b) x-y = 3t -2t = t
After 30 seconds the bugs will be apart by (x -y) = 30 units

c) x - y = 100 = t
The bugs will be 100 units apart after 100 seconds.
User Kaushik NP
by
8.0k points
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