A baseball diamond is a square with side 90 feet. A batter hits the ball and runs toward first base with a speed of 24 f/s.
At what rate is his distance from second base changing when he is halfway to first base?
Well, it seems our batter is quite the sprinter! Now, to tackle this problem, let's break it down.
First, we need to find the distance from our batter to second base when he is halfway to first base. That would be half the length of a side of the square, which is 45 feet.
Now, let's find the rate at which his distance from second base is changing. We can use a bit of geometry here. Since the bases form a square, the distance between the batter and second base is equal to the square root of (45^2 + 90^2).
But let's not get all square-eyed with the numbers. So, we'll simplify things a bit. 45 squared is 2025, and 90 squared is 8100.
So, the distance from the batter to second base is equal to the square root of (2025 + 8100), which is the square root of 10125.
Now, let's calculate this funny business. The square root of 10125 is approximately 100.6 feet.
Therefore, the rate at which the batter's distance from second base is changing when he is halfway to first base is approximately 100.6 f/s.
I hope that answers your question and keeps you entertained along the way!
To find the rate at which the distance from second base is changing when the batter is halfway to first base, we need to use the concept of related rates.
Let's assume the distance from the batter to first base is x feet. We know the baseball diamond is a square with sides measuring 90 feet, so the distance from second base to the halfway point between first base and home plate is also x feet. Let's call this distance y.
Since the batter is running towards first base, the distance between the batter and second base is decreasing over time, and we want to find the rate of change of that distance, i.e., dx/dt.
We can form a right-angled triangle using the distances x, y, and the side of the square (90 feet).
Using the Pythagorean theorem, we can write:
x^2 + y^2 = (90)^2
Differentiating with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we're interested in finding dx/dt when the batter is halfway to first base, we can substitute y = x/2 into the equation above:
2x(dx/dt) + 2(x/2)(dy/dt) = 0
Simplifying further:
2x(dx/dt) + x(dy/dt) = 0
Factoring out x:
x(2(dx/dt) + (dy/dt)) = 0
Since we're interested in dx/dt, we can rearrange the equation:
2(dx/dt) + (dy/dt) = 0
Substituting the given speed of the batter, dy/dt = 24 ft/s:
2(dx/dt) + 24 = 0
Rearranging the equation to solve for dx/dt:
2(dx/dt) = -24
dx/dt = -24/2
dx/dt = -12 ft/s
Therefore, the rate at which the distance from second base is changing when the batter is halfway to first base is -12 ft/s. The negative sign indicates that the distance is decreasing.
To find the rate at which the batter's distance from second base is changing, we need to determine his position as a function of time and then differentiate that function with respect to time.
First, let's draw a diagram to visualize the situation. Assume that the diamond is positioned with second base at the origin (0, 0) and the batter starting at point (0, -90) and running towards first base along the positive y-axis.
Let's consider time (t) as the independent variable. The position of the batter at any time t can be described by the coordinates (0, -90 + 24t), where the first element represents the x-coordinate and the second element represents the y-coordinate.
Since we want to find the rate at which the batter's distance from second base is changing when he is halfway to first base, we need to find his position at that moment.
The halfway point to first base can be represented by the coordinates (0, -45). So we need to find the time t when the y-coordinate is equal to -45.
-90 + 24t = -45
24t = -45 + 90
24t = 45
t = 45 / 24
t = 1.875 seconds
Now that we have determined the time when the batter is halfway to first base, we can find the rate of change of his distance from second base.
To do this, we need to find the derivative of the distance formula. The distance (d) between two points (x1, y1) and (x2, y2) can be calculated using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, we want to find the rate of change of the distance between the batter's position (0, -45) and second base (0, 0) with respect to time (t).
So, the distance formula becomes:
d = sqrt((0 - 0)^2 + (-45 - 0)^2)
= sqrt(0 + 2025)
= 45 feet
Now, we can differentiate the distance formula with respect to time to find the rate of change:
dd/dt = d(sqrt((x2 - x1)^2 + (y2 - y1)^2)) / dt
Since x2, y2, x1, and y1 are all constants in this case, we can simplify the equation to:
dd/dt = (d/dt)sqrt(0 + 2025)
= (d/dt)sqrt(2025)
= (1/2sqrt(2025))(d/dt)(2025)
= (1/2sqrt(2025))(0)
= 0
Therefore, the rate at which the batter's distance from second base is changing when he is halfway to first base is 0 ft/s.