Position on graph

Question

Kirby moves with constant speed 5 units per hour along the line y=3x/4 +6, crossing
the y-axis at midnight and the x-axis later. When is the x-axis crossing made? What does it
mean to say that Kirby’s position is a function of time? What is Kirby’s position 1.5 hours
after midnight? What is Kirby’s position t hours after midnight? 

Answer 1

y = 3x/4 + 6

to find y-intercept set x=0

y-intercept: (0,6) = midnight

 

to find x-intercept set y=0

0 = 3x/4 + 6

3x/4 = -6

x = -6(4/3)

x = -8

x-intercept: (-8,0)

 

The distance from midnight (0,6)

and x-intercept (-8,0) is

d = √[(-8-0)²+(0-6)²]

   = √100

   = 10

 

the distance from midnight to crossing

x-axis is 10 units.  Moving at 5 units per

hour this took 2 hours to traverse.

Thus Kirby crosses the x-axis at 2 hours

after midnight or 2 a.m.

 

1 1/2 hours after midnight at 5 units per hour

is a distance of 5(1.5) units or 7.5 units from

(0,6).  The midpoint of (0,6) and (-8,0) is

[(0-8)/2 , (6-0)/2] = (-4,3) ... This is 5 units

from each of the y-intercept and the x-intercept.

Going 2.5 more units in the next half hour will

put Kirby at the midpoint of (-4,3) and the

x-intercept (-8,0).

 

Location at 1 1/2 hours after midnight is:

[(-4-8)/2, (3-0)/2] = (-6, 3/2)

 

Picture a (slow) snail crawling along a piece of rope

tied at an angle between a low spot and a high spot..

The snail maintains a steady pace of 5 inches

per hour along this path... His position is a function of

time just as the position of a car travelling a straight

road at a steady speed from city A to city B would be a function of time elapsed.

 

*********************************

The location at time t hours is a distance

of 5t units from (0,6) noting that the line

is being traversed "backwards" since the

x-values are getting larger in the negative

direction as time passes...

 

Let (x_t, y_t) denote the position at t hours

after midnight.

 

5t = √[(x_t-0)²+(y_t-6)²]

5t = √[(x_t)² + (y_t-6)²]

25t² = x_t² + (y_t -6)²

25t² = x_t² + y_t²-12y_t + 36

 

Since this is on the line y = 3x/4+6

we have:

 

25t² = x_t² + (3x_t/4 + 6)² -12(3x_t/4 + 6) + 36

25t² = x_t² + 9x_t²/16 + 9x_t + 36 - 9x_t - 72 + 36

25t² = x_t² + (9x_t²)/16

multiply through by 16

400t² = 16x_t² + 9x_t²

400t² = 25x_t²

16t² = x_t²

 

From this we can plug in a value for t

and solve for x_t.   To find y_t use the

equation y_t = 3x/4 + 6

 

Noting that the x_t will be a negative value

and once we pass (-8,0) the y_t value will

also be negative.

 

Example:  We know the midpoint of

(0,6) and (-8,0) is (-4,3) and this occurred

at time t=1 hour after midnight.

 

16(1)² = x_t²

16 = x_t²

x_t = ±4

Since we know the x value is negative

we have x_t = -4

 

y_t = 3x_t/4 + 6 = -3 +6 = 3

The point is (-4,3) which coincides with

what we already know to be true.