Answer:
C. Ari and Matthew collide at 4.8 seconds.
Explanation:
Ari and Matthew will collide when they have the same x and y position. Since Ari's path is given by
x(t) = 36 + (1/6)t
y(t) = 24 + (1/8)t
And Matthew's path is given by
x(t) = 32 + (1/4)t
y(t) = 18 + (1/4)t
We need to make x(t) equal for both, so we need to solve the following equation
Ari's x(t) = Matthew's x(t)
36 + (1/6)t = 32 + (1/4)t
Solving for t, we get
36 + (1/6)t - (1/6)t = 32 + (1/4)t - (1/6)t
36 = 32 + (1/12)t
36 - 32 = 32 + (1/12)t - 32
4 = (1/12)t
12(4) = 12(1/12)t
48 = t
It means that after 48 tenths of seconds, Ari and Mattew have the same x-position. To know if they have the same y-position, we need to replace t = 48 on both equations for y(t)
Ari's y position
y(t) = 24 + (1/8)t
y(t) = 24 + (1/8)(48)
y(t) = 24 + 6
y(t) = 30
Matthew's y position
y(t) = 18 + (1/4)t
y(t) = 18 + (1/4)(48)
y(t) = 18 + 12
y(t) = 30
Therefore, at 48 tenths of a second, Ari and Mattew have the same x and y position. So, the answer is
C. Ari and Matthew collide at 4.8 seconds.
