Observing Bug on Paper

Question

I have been observing a bug walking across my graph paper. When I first started watching it it was at (-1,12), going at a consistent speed and direction it passes (2,8) 2 seconds later.

a. If it continues in this direction when will it cross the x-axis?

b.If it continues at this speed where will it be 10 seconds from the starting position?

c. How much time does the bug spend in the first quadrant?

Answer 1

To answer these questions, it is handy to find the equation of the line. For simplicity's sake, let us get this into slope-intercept form (y = mx+ b). To do so, I will first need to find the slope of the line.

Slope can be found by taking rise over run in a single direction, so lets try left to right. Rise = 12-8=4 (difference in y coordinates), run = (-1)- 2 = -3, so the slope must be -4/3. This yields that y = (-4/3)x +b. Then, lets plug in a known (x,y) pair on this line, we have 2 given points to select from.

y = (-4/3)x +b --> 12 = (-4/3)(-1) + b --> 12 = 4/3 + b --> 32/3 = b --> y = (-4/3)x + (32/3)

a. Since the slope is not zero, this line must cross the x-axis at some point in space.

b. For this one, we can use a trick to answer the question. We should note that we have a change in position already from start to 2 seconds later. In 2 seconds, the bug moved 3 right (2 - -1), and down 4 (12 - 8). At the same speed, we would thus repeat this motion every 2 seconds. So to get from start to end (10 seconds), we would move 5x the amount moved in 2 seconds. Thus, right 15 (3x5), and down 20 (4x5). Starting from (-1,12), we would end up at (14, -8).

c. To do this, we need to find 2 things: 1) the speed of the bug , and 2) the points where this bug enters and leaves the first quadrant (where it hits both the positive x and positive y axes).

Speed is distance over time, so lets find the distance the bug covers in the 2 second interval we are given.

Distance = sqrt[ (12-8)^2 + (-1-2)^2] = sqrt[4^2 + 3^2] = sqrt [16+9] = sqrt[25] = 5. The bug's speed is 5 units every 2 seconds, or 2.5 units per second.

We have the y-interecept (32/3) from our initial equation of the line, so now we just need to find the x-intercept. To do so, set y to 0 and solve for x.

0= (-4/3)(x) + 32/3 --> (4/3)x = 32/3 --> 4x = 32 --> x = 8. Thus the x-intercept is (8,0). We see that it hits both positive x and y axes, thus there is a limited time the bug is in the first quadrant. Final step is to set up a ratio showing that the speed is constant.

5 units / 2 seconds = sqrt[(8-0)^2)+(0-32/3)^2] units / x seconds, where x is the time the bug spends in that interval

-->5 / 2 = sqrt[64+1124/9] / x

--> 5 / 2 = sqrt [576/9 + 1124/9] / x

--> 5 / 2 = sqrt [1700/9] / x

--> 5 / 2 = 13.7437 / x

cross mulitiply --> 5x = 57.487 --> x = 5.497 seconds, so the bug is in the 1st quadrant for ~5.5 seconds.

If it only hits one of these 2 positive axes, we have an infinite time in the 1st quadrant and if we hit neither of these axes then the bug spends no time in the first quadrant.