Question

A robot, always moving at a constant speed of 2 units per second, starts at position (20,50) on the coordinate plane

and heads in a southeast direction along the line 3x + 4y = 260. After 15 seconds, it turns clockwise 90° and
travels in a straight line in this new direction.
a. What are the coordinates of the point at which the robot made the turn? What might be a relatively
straightforward way of determining this point?
b. Find an equation for the second line on which the robot traveled.
c. If, after turning, the robot travels for 20 seconds along this line and then stops, how far will it be from its
starting position?
d. What is the equation of the line the robot needs to travel along in order to now return to its starting position?
How long will it take for the robot to get there?

Answer 1

a) (44,32) are the coordinates. This can be found by finding the slope of the line and writing the distance travelled in function of how much did the robot move on the x-axis (note that you can obtain how much the robot moved because you get the speed and the travel time)

b) An equation for the second line is 4x -3y = 80

c) It will be 50 units far

d) The equation of the line is x=20. It takes 25 seconds for the robot to reach its starting position.

Explanation:

After 15 seconds, the robot moved 2*15 = 30 units.

The line 3x + 4y = 260 can be written this way

4y = 260-3x

y = -3/4 x + 65

Hence its slope is -3/4

If the robot advances k units in the x-coordinate, then it should advance -3/4 k in the y-coordinate, hence we need k such that

k² + (-3/4k)² = 30² = 900

Therefore 25/16 k² = 900, k² = 575

so k = √576 = 24

As a result, after 15 seconds, the robot moved 24 units in the x-axis and -3/4*24 = -18 units on the y-axis (18 untis down).

Since it started in (20,50), then it ends in (44,32), that is the moment where the robot turned.

If it turns 90º clockwise, then it should go backwars in the x-coordinate.

The slope of the perpendicular line is 4/3 (because the product is -1)

Also, for x = 44, y = 32, therefore

4(x-44) - 3(y-32) = 0

Or, equivalently

4x -3y = 80

After 20 seconds it moved 20*2 = 40 units.

Since the slope this time is 4/3, it moved (k, 4/3k) for certain value of k

Then

k² + (4/3 k)² = 40² = 1600

Thus,

25/9 k² = 1600,

hence k = - √(1600/(25/9)) = -24 (it takes a negative value because it goes backwards)

If k = -24, then it moved -24 *4/3 = -32 in the y-coordinate

The end position is, therefore, (44- 24, 32 - 32) = (20,0)

The distance between (20,0) and the original position (20,50) is, as a consecuence, 50.

Therefore, if the robot moves in a straight line to return, it will take 50/2 = 25 seconds.

The robot returns in a vertical path starting in (20,0) and ending in (20,50)

The equation of the line is x = 20