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Chapter 2: Problem 6

For the train of Example \(2.2,\) find the average velocity between 3: 14 P.M. when the train is at \(3 \mathrm{km}\) east of the origin and 3: 28 P.M. when it is \(10 \mathrm{km}\) east of the origin.

Short Answer

Expert verified
Answer: The average velocity of the train between 3:14 PM and 3:28 PM is 30 km/h east.

Step by step solution

01

Determine the displacement

The displacement is the change in position between the two times. We are given the position at 3:14 PM (3 km east of the origin) and the position at 3:28 PM (10 km east of the origin). To find the displacement, subtract the initial position from the final position: Displacement = Final position - Initial position.
02

Find the change in time

We are given the times 3:14 PM and 3:28 PM, so we need to find the difference between these two times. Subtract the initial time from the final time: Change in time = Final time - Initial time.
03

Calculate the average velocity

To find the average velocity, we need to divide the displacement by the change in time: Average Velocity = Displacement / Change in time. Now let's calculate the values.
04

Calculate the displacement

Displacement = 10 km (final position) - 3 km (initial position) = 7 km (east)
05

Calculate the change in time

Change in time = 3:28 PM - 3:14 PM = 14 minutes (Note that we need to convert the time to the same unit as the displacement, which is in hours). So, Change in time = 14 minutes * (1 hour / 60 minutes) = 7/30 hours
06

Calculate the average velocity

Average Velocity = Displacement / Change in time = (7 km) / (7/30 hours) = (7 km) * (30/7 hours) = 30 km/h (east) So, the average velocity of the train between 3:14 PM and 3:28 PM is 30 km/h east.

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