/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 In a relay race, each contestant... [FREE SOLUTION] | 91Ó°ÊÓ

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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 m in 20.0 s. On the return trip she is more confident and takes only 15.0 s. What is the magnitude of her average velocity for (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

Understand Average Velocity

Average velocity is defined as the total displacement divided by the total time taken. To calculate the average velocity for each segment, we need to determine the displacement and the time for that segment.
02

Calculate Average Velocity for First 25.0 m

For the first 25.0 m, Edith's displacement is 25.0 m and the time taken is 20.0 seconds. The average velocity is calculated as \( v_1 = \frac{\text{displacement}}{\text{time}} = \frac{25.0 \text{ m}}{20.0 \text{ s}} = 1.25 \text{ m/s} \).
03

Calculate Average Velocity for Return Trip

On the return trip, the displacement is again 25.0 m but in the opposite direction, and the time taken is 15.0 seconds. The average velocity is \( v_2 = \frac{25.0 \text{ m}}{15.0 \text{ s}} = 1.67 \text{ m/s} \).
04

Calculate Average Velocity for Entire Round Trip

The displacement for the entire round trip is zero because she returns to the starting point. Average velocity is \( \frac{\text{total displacement}}{\text{total time}} = \frac{0 \text{ m}}{35.0 \text{ s}} = 0 \text{ m/s} \).
05

Calculate Average Speed for the Round Trip

Average speed is the total distance traveled divided by the total time. The total distance is 50.0 m (25.0 m each way), and the total time is 35.0 seconds. Average speed is \( \frac{50.0 \text{ m}}{35.0 \text{ s}} = 1.43 \text{ m/s} \).

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Key Concepts

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

Average Speed
Average speed is a key concept in understanding motion. It reflects how fast an object moves, disregarding direction. To grasp this, consider Edith's relay race run. In total, she travels 50.0 meters — 25.0 meters out and then 25.0 meters back. The average speed is determined by dividing this total distance by the total time taken for the journey. In Edith's case, she spends 20 seconds on her outward trip and 15 seconds returning, totaling 35 seconds. Hence, her average speed can be calculated as follows:
  • Total distance: 50.0 m
  • Total time: 35.0 s
The formula for average speed is:\[\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{50.0 \text{ m}}{35.0 \text{ s}} = 1.43 \text{ m/s}\]Average speed is always a positive number as it doesn't involve direction, just total distance and time.
Displacement
When discussing motion, 'displacement' points to the change in position of an object. It's a vector quantity, which means it has both a magnitude and a direction. This is different from 'distance,' which only measures the path traveled, not the direction. In the relay race scenario, Edith's displacement for each segment: the initial 25.0 m and the 25.0 m return trip, is straightforward. Initially, her displacement is exactly 25.0 m towards her target. However, when she completes the round trip by returning to the starting point, the net displacement becomes zero. Displacement focuses on the change in position from the initial to the final point, which can be given as:
  • Initial displacement for the first 25.0 m: 25.0 m
  • Displacement returning to start: -25.0 m
  • Total displacement for round trip: 0 m
Displacement can sometimes be tricky, especially when motion involves backtracking to the starting point.
Kinematics
Kinematics is the branch of physics describing the motion of objects without considering the forces causing them. It provides the tools to describe motion through concepts like velocity, speed, and displacement. In Edith's race, the kinematics principles help determine specifics like her average velocities and speeds at different parts of her run. Her motion can be quantified by:
  • Velocity, which incorporates both speed and direction, making it a vector.
  • Speed, which is scalar, representing only how fast she's moving regardless of direction.
Kinematics involves examining Edith's average velocities. For instance, running the first 25.0 m at 1.25 m/s and the returning 25.0 m at 1.67 m/s reflects the application of kinematics concepts, giving clear insights into the nature of her motion.
Round Trip Calculation
Calculating motion for a round trip involves special attention since the starting and ending points are the same. In such scenarios, one might compute both average speed and average velocity.For Edith's round trip in the relay:
  • The total distance covered is 50.0 m, considering her 25.0 m outbound and 25.0 m return leg.
  • Average velocity differs here because her starting and ending points are the same, which makes the displacement zero.
Displacement being zero impacts the average velocity calculation, since:\[\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{0 \text{ m}}{35.0 \text{ s}} = 0 \text{ m/s}\]In contrast, the average speed considers only the total distance and time, giving a clearer perspective on Edith's progress throughout her race.

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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 direction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2}\) (a) Find the turtle'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 \(t=40 \mathrm{s}.\)

A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time sled is 14.4 \(\mathrm{m}\) from the top, 2.00 s after that it is 25.6 \(\mathrm{m}\) from the top, 2.00 s later 40.0 \(\mathrm{m}\) from the top, and 2.00 s later it is 57.6 \(\mathrm{m}\) from the top. (a) What is the magnitude of the average velocity of the sled during each of the 2.00 -s intervals after passing the \(14.4-\mathrm{m}\) point? (b) What is the acceleration of the sled? (c) What is the speed of the sled when it passes the \(14.4-\mathrm{m}\) point? (d) How much time did it take to go from the top to the \(14.4-\mathrm{m}\) point? (e) How far did the sled go during the first second after passing the \(14.4-\mathrm{m}\) point?

Mars Landing. In January \(2004,\) NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages: Stage \(A:\) Friction with the atmosphere reduced the speed from \(19,300 \mathrm{km} / \mathrm{h}\) to 1600 \(\mathrm{km} / \mathrm{h}\) in 4.0 \(\mathrm{min.}\) Stage \(B:\) A parachute then opened to slow it down to 321 \(\mathrm{km} / \mathrm{h}\) in 94 \(\mathrm{s} .\) Stage \(C :\) Retro rockets then fired to reduce its speed to zero over a distance of 75 \(\mathrm{m}.\) Assume that each stage followed immediately after the preceding one and that the acceleration during each stage was constant. (a) Find the rocket's acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) during each stage. (b) What total distance (in km) did the rocket travel during stages \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C} ?\)

An antelope moving with constant acceleration covers the distance between two points 70.0 \(\mathrm{m}\) apart in 7.00 \(\mathrm{s} .\) Its speed as it passes the second point is 15.0 \(\mathrm{m} / \mathrm{s}\) . (a) What is its speed at the first point? (b) What is its acceleration?

Two cars, \(A\) and \(B,\) travel in a straight line. The dis tance of \(A\) from the starting point is given as a function of time by \(x_{A}(t)=\alpha t+\beta t^{2},\) with \(\alpha=2.60 \mathrm{m} / \mathrm{s}\) and \(\beta=1.20 \mathrm{m} / \mathrm{s}^{2} .\) The distance of \(B\) from the starting point is \(x_{B}(t)=\gamma t^{2}-\delta t^{3},\) with \(\gamma=2.80 \mathrm{m} / \mathrm{s}^{2}\) and \(\delta=0.20 \mathrm{m} / \mathrm{s}^{3} .\) (a) Which car is ahead just after they leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

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