Question 2.3: CATCHING A FLY BALL GOAL Apply the definition of instantaneo...
CATCHING A FLY BALL
GOAL Apply the definition of instantaneous acceleration.
PROBLEM A baseball player moves in a straight line path in order to catch a fly ball hit to the outfield. His velocity as a func tion of time is shown in Figure 2.11a. Find his instantaneous acceleration at points Ⓐ, Ⓑ, and ©.
STRATEGY At each point, the velocity vs. time graph is a straight line segment, so the instantaneous acceleration will be the slope of that segment. Select two points on each segment and use them to calculate the slope.
Acceleration at Ⓐ.
The acceleration at Ⓐ equals the slope of the line connecting the points (0 \mathrm{~s}, 0 \mathrm{~m} / \mathrm{s}) and (2.0 \mathrm{~s}, 4.0 \mathrm{~m} / \mathrm{s}) :
a=\frac{\Delta v}{\Delta t}=\frac{4.0 \mathrm{~m} / \mathrm{s} -0}{2.0 \mathrm{~s}-0}=+2.0 \mathrm{~m} / \mathrm{s}^{2}
Acceleration at Ⓑ.
\Delta v=0, because the segment is horizontal:
a=\frac{\Delta v}{\Delta t}=\frac{4.0 \mathrm{~m} / \mathrm{s}-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}-2.0 \mathrm{~s}}=0 \mathrm{~m} / \mathrm{s}^{2}
Acceleration at ©.
The acceleration at © equals the slope of the line connecting the points (3.0 \mathrm{~s}, 4.0 \mathrm{~m} / \mathrm{s}) and (4.0 \mathrm{~s}, 2.0 \mathrm{~m} / \mathrm{s}) :
a=\frac{\Delta v}{\Delta t}=\frac{2.0 \mathrm{~m} / \mathrm{s}-4.0 \mathrm{~m} / \mathrm{s}}{4.0 \mathrm{~s}-3.0 \mathrm{~s}}=-2.0 \mathrm{~m} / \mathrm{s}^{2}
REMARKS For the first 2.0 \mathrm{~s}, the ballplayer moves in the positive x-direction (the velocity is positive) and steadily accelerates (the curve is steadily rising) to a maximum speed of 4.0 \mathrm{~m} / \mathrm{s}. He moves for 1.0 \mathrm{~s} at a steady speed of 4.0 \mathrm{~m} / \mathrm{s} and then slows down in the last second (the v vs. t curve is falling), still moving in the positive x-direction ( v is always positive).