A baseball player hits a baseball from home base towards second base with a velocity of 90 feet/second. At that same moment, he runs from home base towards first base at 30 feet/second. Find the rate of change of the distance between him and the baseball after 1 second.
so, how do you figure the distance? Look up the various measurements of a ball field.
Then just take the derivatives and plug in your numbers.
To find the rate of change of the distance between the baseball player and the baseball after 1 second, we can consider the positions of both the baseball player and the baseball at two different times: initially (t = 0) and 1 second later (t = 1).
Let's denote the distance between the baseball player and the baseball at time t as D(t). Initially, when t = 0, the distance D(0) between them is the distance from home base to second base, which is 90 feet.
One second later, when t = 1, the baseball player runs towards first base at a velocity of 30 feet/second, so he moves a distance of 30 feet towards first base. The baseball, on the other hand, was hit from home base towards second base with a velocity of 90 feet/second. In one second, it would have traveled a distance of 90 feet towards second base.
Now, we can calculate the new distance D(1) between the baseball player and the baseball:
D(1) = D(0) - (distance baseball player moved) + (distance baseball moved)
D(1) = 90 ft - 30 ft + 90 ft
D(1) = 150 ft
Therefore, after 1 second, the distance between the baseball player and the baseball is 150 feet.
To find the rate of change of the distance between them after 1 second, we can calculate the derivative of D(t) with respect to t at t = 1. This will give us the rate of change, or the velocity, at that specific moment. Since the distances traveled are constant, the derivative is simply 0.
Hence, the rate of change of the distance between the baseball player and the baseball after 1 second is 0 feet/second.