/*! 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 15 A runner accelerates to a veloci... [FREE SOLUTION] | 91Ó°ÊÓ

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A runner accelerates to a velocity of \(4.15 \mathrm{~m} / \mathrm{s}\) due west in \(1.50 \mathrm{~s} .\) His average acceleration is \(0.640 \mathrm{~m} / \mathrm{s}^{2},\) also directed due west. What was his velocity when he began accelerating?

Short Answer

Expert verified
The initial velocity was 3.2 m/s due west.

Step by step solution

01

Understand the Given Values

We need to identify the values given in the problem. We have:- Final velocity, \(v = 4.15 \, \text{m/s}\) (due west)- Time, \(t = 1.50 \, \text{s}\)- Average acceleration, \(a = 0.640 \, \text{m/s}^2\) (due west).

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Key Concepts

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

Acceleration
Acceleration refers to how quickly an object’s velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. In the context of kinematics, understanding acceleration is key to solving problems where an object’s speed changes.
For our problem, the average acceleration is given as 0.640 m/s² directed due west. This tells us how much the runner's velocity increases per second in a specific direction.
When calculating acceleration, we often use the formula: \[ a = \frac{\Delta v}{t} \]where \(a\) is the average acceleration, \(\Delta v\) is the change in velocity, and \(t\) is the time over which the change occurs. Since acceleration here is positive, this implies that the object is speeding up in the direction specified, which is due west in this case.
Velocity
Velocity is a measure of both the speed and direction of an object’s motion. Unlike speed, velocity is a vector, meaning it accounts for direction. This distinction is crucial for solving kinematics problems in physics.
In the original exercise, we are told that the final velocity of the runner is 4.15 m/s due west. This velocity indicates not only how fast the runner is moving but also the direction of the movement.
To find the initial velocity when the runner began accelerating, we can rearrange the equation for acceleration:\[ v_i = v_f - (a \times t) \] where \(v_i\) is the initial velocity, \(v_f\) is the final velocity, \(a\) is acceleration, and \(t\) is time. Using this formula, students can plug in the numbers to find how fast the runner was moving at the start of the acceleration.
Time
Time is an integral component in kinematics, representing the duration over which motion occurs or changes. In problems involving kinematics, time helps determine how other variables like velocity and acceleration interact.
In the given problem, the time span during which the runner accelerates is 1.50 seconds. Knowing the duration of this phase allows us to calculate the change in velocity with given acceleration.
Time, denoted by \(t\), is the period during which all these changes are measured. In equations, remember to keep the units consistent; here, time is measured in seconds. Proper understanding of how time fits into kinematic equations enables students to link various aspects of motion effectively. During calculations, accurately accounting for time is crucial, as even slight errors in the time measure can significantly affect the outcome of kinematic equations.

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Most popular questions from this chapter

At provides some background for this problem. A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. (a) Does the acceleration of the ball reverse direction when the ball starts its downward trip? (b) What is the displacement of the ball when it returns to its point of release? Explain your answers. Problem If the ball is in the air for a total time of \(8.0 \mathrm{~s}\), what is its initial velocity?

A football player, starting from rest at the line of scrimmage, accelerates along a straight line for a time of \(1.5 \mathrm{~s}\). Then, during a negligible amount of time, he changes the magnitude of his acceleration to a value of \(1.1 \mathrm{~m} / \mathrm{s}^{2}\). With this acceleration, he continues in the same direction for another \(1.2 \mathrm{~s},\) until he reaches a speed of \(3.4 \mathrm{~m} / \mathrm{s} .\) What is the value of his acceleration (assumed to be constant) during the initial 1.5 -s period?

Interactive Solution \(\underline{2.31}\) at offers help in modeling this problem. A car is traveling at a constant speed of \(33 \mathrm{~m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{~km}\) away?

A jogger accelerates from rest to \(3.0 \mathrm{~m} / \mathrm{s}\) in \(2.0 \mathrm{~s}\). A car accelerates from 38.0 to \(41.0 \mathrm{~m} / \mathrm{s}\) also in \(2.0 \mathrm{~s}\). (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the \(2.0 \mathrm{~s}\) ? If so, how much farther?

(a) The acceleration of a NASCAR race car is zero. Does this necessarily mean that the velocity of the car is also zero? (b) If the speed of the car is constant as it goes around the track, is its average acceleration necessarily zero? Justify your answers. Problem (a) Suppose that the race car is moving to the right with a constant velocity of \(+82 \mathrm{~m} / \mathrm{s}\). What is the average acceleration of the car? (b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car? Verify that your answers agree with your answers to the Concept Questions.

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