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Chapter 2: Problem 56

Relay Race. In a relay race, each contestant runs 25.0 \(\mathrm{m}\) while carrying an egg balanced on a spoon, turns around, and comes back to the starting point. Edith runs the first 25.0 \(\mathrm{m}\) in 20.0 \(\mathrm{s}\) . On the return trip she is more confident and takes only 15.0 \(\mathrm{s}\) . What is the magnitude of her average velocity for \((\mathrm{a})\) the first 25.0 \(\mathrm{m} ?\) (b) The return trip? (c) What is her average velocity for the entire round trip? (d) What is her average speed for the round trip?

Short Answer

Expert verified
(a) 1.25 m/s, (b) 1.67 m/s, (c) 0 m/s, (d) 1.43 m/s.

Step by step solution

01

Calculate Average Velocity for the First 25.0 m

Average velocity is defined as the total displacement divided by the total time taken. For the first part of her run (25.0 m):1. Displacement, \( s = 25.0 \, \mathrm{m} \)2. Time, \( t = 20.0 \, \mathrm{s} \)3. Average velocity, \( v = \frac{s}{t} = \frac{25.0 \, \mathrm{m}}{20.0 \, \mathrm{s}} = 1.25 \, \mathrm{m/s} \)
02

Calculate Average Velocity for the Return Trip

On the return trip, Edith covers the same distance in a shorter time:1. Displacement, \( s = 25.0 \, \mathrm{m} \)2. Time, \( t = 15.0 \, \mathrm{s} \)3. Average velocity, \( v = \frac{s}{t} = \frac{25.0 \, \mathrm{m}}{15.0 \, \mathrm{s}} = 1.67 \, \mathrm{m/s} \)
03

Calculate Average Velocity for the Entire Round Trip

The total displacement for a round trip is zero because Edith returns to her starting point.1. Displacement for round trip = 0 m2. Total time taken = \( 20.0 \, \mathrm{s} + 15.0 \, \mathrm{s} = 35.0 \, \mathrm{s} \)3. Average velocity, \( v = \frac{0 \, \mathrm{m}}{35.0 \, \mathrm{s}} = 0 \, \mathrm{m/s} \)
04

Calculate Average Speed for the Round Trip

Average speed is the total distance covered divided by the total time taken. Unlike velocity, it's not a vector so it won't be zero even for a round trip.1. Total distance for the round trip = \( 25.0 \, \mathrm{m} + 25.0 \, \mathrm{m} = 50.0 \, \mathrm{m} \)2. Total time = 35.0 s3. Average speed \( = \frac{50.0 \, \mathrm{m}}{35.0 \, \mathrm{s}} \approx 1.43 \, \mathrm{m/s} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Displacement
Displacement is an important concept to grasp, especially when discussing motion and velocity. It refers to the change in position of an object and takes into account only the initial and final positions, not the path taken in between. In other words, displacement is a vector quantity, meaning it has both magnitude and direction.
  • For a single straight run, like moving 25 meters in one direction, displacement equals the actual distance covered, here it is 25 meters.
  • However, with a round trip, where a starting and ending point are the same, displacement becomes zero.
In the relay race example, Edith's displacement for each 25-meter run is simply that 鈥� 25 meters. But because she starts and ends at the same point for the round trip, her overall displacement is zero. This is crucial for understanding how average velocity is affected by directional movement.
Average Speed
Average speed differs from average velocity due to its focus on total distance rather than displacement. While velocity measures how quickly position changes, speed measures how quickly distance is covered.
  • To find average speed, divide the total distance by the total time taken.
  • In a round trip, because the distance traveled is non-zero (Edith ran 25 meters out and 25 meters back, totaling 50 meters), average speed becomes a meaningful measure of her motion.
In the context of Edith's race, her total distance for the trip was 50 meters, covered in 35 seconds, leading to an average speed of approximately 1.43 m/s. This reflects her consistent ability to cover ground over time without regard to starting or ending position.
Distance and Time Calculations
Calculating distance and time effectively is foundational for understanding kinematics. Distance accounts for the entire path traveled, while time is the duration of travel.
  • For calculating average speed, sum up all segments of travel distance. In Edith's case, 25 meters each way equals a total of 50 meters.
  • Time should include every second in motion: Edith's expeditions took a total of 35 seconds.
  • It's essential to remember that both distance and time remain positive; they do not consider direction as displacement does.
Working through this example, Edith's performance showcases a basic application of these principles. Her calculations show how average speed gives insight into how quickly she covered the path, independent from changes in position.

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Most popular questions from this chapter

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}\)

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The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) (about 45 \(\mathrm{mi} / \mathrm{h} )\) . Initially, the car is also traveling at 20.0 \(\mathrm{m} / \mathrm{s}\) and its front bumper is 24.0 \(\mathrm{m}\) behind the truck's rear bumper. The car accelerates at a constant 0.600 \(\mathrm{m} / \mathrm{s}^{2}\) , then pulls back into the truck's lane when the rear of the car is 26.0 \(\mathrm{m}\) ahead of the front of the truck. The car is 4.5 \(\mathrm{m}\) long and the truck is 21.0 \(\mathrm{m}\) long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

Entering the Freeway. A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 \(\mathrm{m} / \mathrm{s}(45 \mathrm{mi} / \mathrm{h})\) when it reaches the end of the 120 \(\mathrm{m}\) -long ramp. \((\mathrm{a})\) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 \(\mathrm{m} / \mathrm{s}\) . What distance does the traffic travel while the car is moving the length of the ramp?
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