At a football game, at the instant the ball goes into play, a player at the line of scrimmage begins running along the edge of the field at a rate of 25 ft/sec. The quarterback receives the ball and is forced into position 30 feet across the field from the boundary line along which the player is running, and 2.5 feet behind the line of scrimmage. 1.5 seconds later, the quarterback is poised to throw the ball. At that instant, what is the rate at which the distance is changing from the quarterback to the player?

Question

At a football game, at the instant the ball goes into play, a player at the line of scrimmage begins running along the edge of the field at a rate of 25 ft/sec. The quarterback receives the ball and is forced into position 30 feet across the field from the boundary line along which the player is running, and 2.5 feet behind the line of scrimmage. 1.5 seconds later, the quarterback is poised to throw the ball. At that instant, what is the rate at which the distance is changing from the quarterback to the player?

Please help I know u need to use the Pythagorean Therom but am not sure how to solve. I know the answer is 20.

Answer 1

let the distance run by the running player be x from the line of scrimmage.

Let the distance between them be d

In my diagram, I have a right-angles triangle with fixed side of 30, the other side (x+2.5) and the hypotenuse d, so that

d^2 = (x+2.5)^2 + 30^2
2d dd/dt = 2(x+2.5) dx/dt
when t = 1.5, x = 37.5 and
d^2 = (40)^2 + 30^2
d = 50
2(50) dd/dt = 2(40)(25)
dd/dt = 80(24)/100 = 19.2 ft/sec

check my arithmetic

Answer 2

shouldnt it be 25 for the last line, not 24? all else looks right tho

Answer 3

To solve this problem, we can use the Pythagorean theorem to find the distance between the quarterback and the player at any given time.

Let's set up a coordinate system with the line of scrimmage as the x-axis and the boundary line as the y-axis. The player's position can be described by the equation:

x = 25t

where x is the player's distance along the x-axis and t is the time in seconds.

The quarterback's position can be described by the equation:

x_qb = 30

y_qb = 2.5

To find the distance between the player and the quarterback, we can use the Pythagorean theorem:

distance^2 = (x - x_qb)^2 + (y - y_qb)^2

Substituting the given values:

distance^2 = (25t - 30)^2 + (0 - 2.5)^2

Simplifying:

distance^2 = (625t^2 - 1500t + 900) + 6.25

distance^2 = 625t^2 - 1500t + 906.25

To find the rate at which the distance is changing, we'll need to take the derivative of the equation with respect to time:

2 * distance * (rate of change of distance) = 1250t - 1500

Substituting the given time, t = 1.5 seconds:

distance * (rate of change of distance) = 1250(1.5) - 1500

rate of change of distance = (1875 - 1500) / distance

Now, we need to find the distance at t = 1.5 seconds:

distance = sqrt(625(1.5)^2 - 1500(1.5) + 906.25)

distance = sqrt(1406.25)

distance ≈ 37.50 ft

Substituting the distance value, we can find the rate of change of distance:

rate of change of distance = (1875 - 1500) / 37.50

rate of change of distance ≈ 40 ft/sec

Therefore, the rate at which the distance is changing from the quarterback to the player at that instant is approximately 40 ft/sec, which is different from the answer you mentioned. Please check your answer or provide further clarification if needed.