A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 26 ft/s.
At what rate is his distance from second base decreasing when he is halfway to first base?
At what rate is his distance from third base increasing at the same moment?
Solution:
Let, a = runner’s distance from second base
x= runner’s distance from first base
Given, x = 26ft/s
dx/dt = runner’s speed
a2 = (90)2 + x2
On differentiating both sides with respect to time, t
2a(da/dt) = 0 + 2x(dx/dt)
a(da/dt) = x(dx/dt)
a(da/dt) = x(26)
When runner is halfway to first base,
a2 = (90)2 + (45)2
a = √10125
a = 100.623
da/dt = (45)(-26) / 100.623
da/dt = -11.62 ft/sec
Therefore, the rate at which the distance is decreasing when the runner is halfway to first base
is -11.62 ft/sec
Let, the distance to third base as a function of the remaining distance to first base = b
b2 = (90)2 + (90 - x)2
On differentiating both sides with respect to time t,
2b db/dt = 2(90 - x) (-1) dx/dt
b db/dt = (90 - x) (-1) dx/dt
db/dt = [(90 - x) (-1) dx/dt]/b
To find the value of b,
b² = (90)² + (90 - 45)²
b² = (90)² + (45)²
b = √10125
b = 100.623
Now, db/dt = [(90 - 45) (-1) (-26)] / 100.623
db/dt = (45) (-1) (-26) / 100.623
db/dt = +11.62 ft/sec
Therefore, the rate at which the distance from third base increasing at the same moment is +11.62 ft/sec.
Therefore, the rate at which the distance is decreasing when the runner is halfway to first base
is -11.62 ft/sec and the rate at which the distance from third base increasing at the same moment is +11.62 ft/sec.
A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 26 ft/s.
Summary:
A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 26 ft/s. The rate at which his distance from second base decreasing when he is halfway to first base is -11.62 ft/sec. The rate at which his distance from third base increasing at the same moment is +11.62 ft/sec.