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Chapter 2: Problem 25

A car travels north at \(30.0 \mathrm{~m} / \mathrm{s}\) for \(10.0 \mathrm{~min}\). It then travels south at \(40.0 \mathrm{~m} / \mathrm{s}\) for \(20.0 \mathrm{~min}\). What are the total distance the car travels and its displacement?

Short Answer

Expert verified
Answer: The total distance traveled is 66,000 meters, and the displacement is 30,000 meters southwards.

Step by step solution

01

Convert time to seconds

To maintain consistent units, convert the given time (in minutes) to seconds by multiplying the time by 60. For the north trip: \(10.0 \mathrm{min} \times 60 \mathrm{s/min} = 600\mathrm{s}\) For the south trip: \(20.0 \mathrm{min} \times 60 \mathrm{s/min} = 1200\mathrm{s}\)
02

Calculate the distance traveled in each direction

To do this, multiply the speed by the time for each direction (north and south): For the north trip: distance = speed \(\times\) time \(= 30.0\mathrm{m/s} \times 600\mathrm{s} = 18000\mathrm{m}\) For the south trip: distance = speed \(\times\) time \(= 40.0\mathrm{m/s} \times 1200\mathrm{s} = 48000\mathrm{m}\)
03

Calculate the total distance traveled

Add up the distances covered in both directions (north and south). Total distance = distance north + distance south \(= 18000\mathrm{m} + 48000\mathrm{m}= 66000\mathrm{m}\)
04

Calculate the displacement

Since the car traveled in opposite directions (north and south), we need to find the difference between the two distances: Displacement = distance south - distance north \(= 48000\mathrm{m} - 18000\mathrm{m} = 30000 \mathrm{m}\) (southwards) Now, we have the total distance traveled and the displacement: Total distance = \(66000\mathrm{m}\) Displacement = \(30000\mathrm{m}\) (southwards)

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Most popular questions from this chapter

The 2007 world record for the men's 100 -m dash was \(9.77 \mathrm{~s}\). The third-place runner crossed the finish line in \(10.07 \mathrm{~s}\). When the winner crossed the finish line, how far was the third-place runner behind him? a) Compute an answer that assumes that each runner ran at his average speed for the entire race. b) Compute another answer that uses the result of Example 2.3, that a world- class sprinter runs at a speed of \(12 \mathrm{~m} / \mathrm{s}\) after an initial acceleration phase. If both runners in this race reach this speed, how far behind is the third-place runner when the winner finishes?

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You are driving at \(29.1 \mathrm{~m} / \mathrm{s}\) when the truck ahead of you comes to a halt \(200.0 \mathrm{~m}\) away from your bumper. Your brakes are in poor condition and you decelerate at a constant rate of \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) a) How close do you come to the bumper of the truck? b) How long does it take you to come to a stop?

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