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Chapter 2: Problem 65

In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. At 3: 00 P.M., a bank robber is spotted driving north on I-15 at milepost $126 .\( His speed is \)112.0 \mathrm{mi} / \mathrm{h}$. At 3: 37 P.M., he is spotted at milepost 185 doing 105.0 milh. During this time interval, what are the bank robber's displacement, average velocity, and average acceleration? (Assume a straight highway.)

Short Answer

Expert verified
Answer: The bank robber's displacement is 94950.06 meters north, his average velocity is 42.7642 m/s north, and his average acceleration is approximately -0.00142072 m/s².

Step by step solution

01

Convert units to meters and seconds

We need to convert the positions and velocities from miles/hour to meters/second. 1 mile = 1609.34 meters 1 hour = 3600 seconds Initial position: 126 miles * 1609.34 m/mile = 202778.84 meters Final position: 185 miles * 1609.34 m/mile = 297728.90 meters Initial velocity: 112.0 mph * (1609.34 m/mile) / (3600 s/hr) = 50.0268 m/s Final velocity: 105.0 mph * (1609.34 m/mile) / (3600 s/hr) = 46.8716 m/s
02

Calculate the time interval

The time interval is given to be from 3:00 P.M. to 3:37 P.M. (37 minutes * 60 seconds/minute) = 2220 s
03

Calculate the displacement

Displacement = Final position - Initial position = 297728.90 m - 202778.84 m = 94950.06 m
04

Calculate the average velocity

Average velocity = Displacement / Time interval = 94950.06 m / 2220 s = 42.7642 m/s (north)
05

Calculate the average acceleration

Average acceleration = (Final velocity - Initial velocity) / Time interval = (46.8716 m/s - 50.0268 m/s) / 2220 s = -0.00142072 m/s² The bank robber's displacement is 94950.06 meters north, his average velocity is 42.7642 m/s north, and his average acceleration is approximately -0.00142072 m/s².

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