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# Math PLEASE HELP! Jo is moving linearly in the xy-plane at a constant speed. She starts from the point (3, 1)?

Jo is moving linearly in the xy-plane at a constant speed. She starts from the point (3, 1)

and moves along the line y = −3x + 10 at a speed of 2 units per second, heading toward

the y-axis.

(a) Write parametric equations for Jo’s location t seconds after starting.

(b) Express Jo’s distance from the point (5, 8) as a function of t.

Please so frustrated give good explanation I need to understand how to do this in general. Thank you

### 2 Answers

- Brach ZLv 41 decade agoFavorite Answer
Jo is moving linearly in the xy-plane at a constant speed. She starts from the point (3,1) and moves along the line y = −3x + 10 at a speed of 2 units per second, heading toward the y-axis.

(a) Write parametric equations for Jo’s location t seconds after starting.

In other words, express her horizontal location, x, and vertical location, y, as functions of time t.

She starts at (3,1) and moves to (x,y) while heading left and up toward the y-axis.

The horizontal distance traveled moving left is 3-x. The vertical distance traveled moving up is y-1.

Since distance/time = speed = 2 units/sec., we can write

(3-x)/t = 2 => 3-x = 2t => x = 3 - 2t.

(y-1)/t = 2 => y-1 = 2t => y = 2t + 1.

(b) Express Jo’s distance from the point (5, 8) as a function of t.

At time t, her location is (x,y). The distance between (x,y) and (5,8) is

√((x-5)² + (y-8)²)

= √((3-2t-5)² + (2t+1-8)²)

= √((-2t-2)² + (2t-7)²)

= √((4t²+8t+4) + (4t²-28t+49))

= √(8t²-20t+53).

NOTE: Since we are told that she starts at (3,1), it follows that at her start time t=0, we must have x=3 and y=1. Therefore, the Haji Koeu's answer cannot be correct.

- Anonymous1 decade ago
a)

simple.

t will be our variable for the amount of seconds. therefore, jo's location is:

x = -t (we use -t, since jo is moving backwards.)

y = -3t + 10

that means, in t seconds, jo's location will be (-t, -3t + 10) [cuz we're plugging in from (x,y)].

b) the distance formula from point (a,b) to point (c,d) is sqrt((a-c)^2 + (b-d)^2). basically, the pythagorean theorem. the location of jo is (-t, -3t + 10), for t seconds. we plug this in against the other point, (5,8) and we get:

sqrt((-t-5)^2 + (-3t + 10 - 8)^2)

which equals:

sqrt((t^2+10t+25) + (9t^2 - 12t + 4)) ..

equaling:

sqrt(10t^2 - 2t + 29)

so sqrt(10t^2 - 2t + 29) is the function that gives us a function of t that gives us the distance from point (5,8) from jo's location..

f(t) = sqrt(10t^2 - 2t + 29)