Question

Ari and Matthew are both running toward the soccer ball during a game.Ari's path is represented by the parametric equations x(t)=36+1/6t,y(t)=24+1/8t, where t is on the interval [0,50] and t is measured in tenths of seconds.Matthew's path is represented by the parametric equations x(t)=32+1/4t,y(t)=18+1/4t, where t is on the interval [0,50] and t is measured in tenths of seconds.Do the boys collide? If so, when do they collide?A. Ari and Matthew do not collide.B. Ari and Matthew collide at 3.2 seconds.C. Ari and Matthew collide at 4.8 seconds.D. Ari and Matthew collide at 2.4 seconds.

Answer 1

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.