Asked by Nathan
A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first base with a speed of 21 ft/s. At what rate is his distance from second base changing when he is halfway to first base? At what rate is his distance from third base changing at the same moment?
Answers
Answered by
Steve
If the runner is at distance x from home plate, the distance to 2nd base is
d^2 = (90-x)^2 + 90^2
so,
2d dd/dt = -2(90-x) dx/dt
when he's halfway there, x=45, so
d = √(45^2+90^2) = 45√5
since dx/dt = 21,
45√5 dd/dt = -45*21
dd/dt = -21/√5
3rd base is similarly done
d^2 = (90-x)^2 + 90^2
so,
2d dd/dt = -2(90-x) dx/dt
when he's halfway there, x=45, so
d = √(45^2+90^2) = 45√5
since dx/dt = 21,
45√5 dd/dt = -45*21
dd/dt = -21/√5
3rd base is similarly done
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