In a game of lawn bowls, the object is to bowl a biased ball so that it gets as close as possible to a smaller white ball called a jack. During a game, a player will sometimes bowl a ball quite quickly so that it travels in a straight line in order to displace an opponent’s ‘guard balls’. In a particular game, player X has 2 guard balls close to the jack. The coordinates of the jack are (0, 0) and the coordinates of the guard balls are A (−1, 4/5) and B (−1/2, 57/40). Player Y bowls a ball so that it travels in a straight line toward the jack. The ball is bowled from position S, which has coordinates (–30, 24).
a. Will player Y displace one of the guard balls? If so, which one?
If you plot the points onto an online graph, it doesn't look like any of the guard balls will be displaced if the ball is bowled straight from 'S' to the jack (0,0)
b. Due to bias, the displaced guard ball is knocked so that it begins to travel in a straight line (at right angles to the path found in part a).
Find the equation of the line of the guard ball.
c. Show that guard ball A is initially heading directly toward guard ball B.
d. Given its initial velocity, guard ball A can travel in a straight line for 1 metre before its bias affects it path. Calculate and explain whether guard ball A will collide with guard ball B
This depends on the size of the balls.
Also, there is no way a ball can be hit so it takes off at a right angle.
Fix up the question, and maybe answers can be determined.
at any rate, to determine the equation of the line, its slope is the negative reciprocal of the incoming line; now you have a slope and a point, so use the point-slope form of the line.
a. To determine if player Y will displace one of the guard balls, we need to find the equation of the straight line between S (-30, 24) and the jack (0, 0). If the line intersects with either guard ball, it means the ball will displace that guard ball.
To find the equation of the line, we can use the slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept.
The slope (m) can be calculated as (change in y) / (change in x):
m = (0 - 24) / (0 - (-30)) = -24 / 30 = -4/5
Now that we have the slope, we can substitute it into the equation:
y = (-4/5)x + b
To find the y-intercept (b), we can plug in the coordinates of one of the points (S or the jack) and solve for b. Let's use the jack's coordinates (0, 0):
0 = (-4/5)(0) + b
0 = b
The final equation of the line is:
y = (-4/5)x
Now, let's check if this line intersects with either of the guard balls' locations A (−1, 4/5) and B (−1/2, 57/40). To do this, we substitute the x and y values of each point into the equation and see if the equation holds true. If it does, the line intersects the point.
For guard ball A:
4/5 = (-4/5)(-1) => 4/5 = 4/5 (The equation holds true)
Therefore, the line intersects with guard ball A.
For guard ball B:
57/40 = (-4/5)(-1/2) => 57/40 = 2/5 (The equation does not hold true)
Therefore, the line does not intersect with guard ball B.
So, player Y will displace guard ball A.
b. Now we need to find the equation of the line that the displaced guard ball A follows after being knocked. We know that the displaced guard ball travels in a straight line at right angles to the original path. Since the original path has a slope of -4/5, the displaced guard ball's line will have a slope of 5/4 (the reciprocal of -4/5) to be perpendicular.
Using the point where guard ball A was originally located (-1, 4/5), we can find the equation of the line with the new slope:
y = mx + b
4/5 = (5/4)(-1) + b
4/5 = -5/4 + b
4/5 + 5/4 = b
16/20 + 25/20 = b
41/20 = b
Therefore, the equation of the line followed by the displaced guard ball A is:
y = (5/4)x + 41/20
c. To show that guard ball A is initially heading directly toward guard ball B, we need to find the slope between the two points A (−1, 4/5) and B (−1/2, 57/40), and if the slope is -1, it means the two points are heading directly towards each other.
The slope between two points can be found using:
m = (change in y) / (change in x)
m = (57/40 - 4/5) / (-1/2 - (-1))
m = (57/40 - 4/5) / (-1/2 + 1)
m = (57/40 - 4/5) / (-1/2 + 2/2)
m = (57/40 - 4/5) / (1/2)
m = (57/40 - 4/5) * (2/1)
m = (57/40 - 32/40) * (2/1)
m = 25/40 * 2/1
m = 50/40 = 5/4
Since the slope between the two points A and B is 5/4 and not -1, we can conclude that guard ball A is not initially heading directly towards guard ball B.
d. To determine if guard ball A will collide with guard ball B, we need to calculate the distance between them and compare it to the distance the guard ball can travel before its bias affects its path.
The distance between A (−1, 4/5) and B (−1/2, 57/40) can be found using the distance formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
d = √((-1/2 - (-1))^2 + (57/40 - 4/5)^2)
d = √((1/2)^2 + (57/40 - 8/10)^2)
d = √(1/4 + (57/40 - 8/10)^2)
d = √(1/4 + (57/40 - 4/5)^2)
d = √(1/4 + (57/40 - 32/40)^2)
d = √(1/4 + (25/40)^2)
d = √(1/4 + 625/1600)
d = √(400/1600 + 625/1600)
d = √(1025/1600)
d ≈ 0.8036 meters
Given that guard ball A can travel 1 meter before its bias affects its path, we can see that the distance between A and B (0.8036 meters) is less than 1 meter. Therefore, guard ball A will collide with guard ball B.