/*! 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 13 A car is stopped for a traffic s... [FREE SOLUTION] | 91Ó°ÊÓ

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A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to \(5.20 \mathrm{~m} / \mathrm{s}\) in \(0.832 \mathrm{~s}\). What are (a) the magnitudes of the linear impulse and (b) the average total force experienced by a \(70.0-\mathrm{kg}\) passenger in the car during the time the car accelerates?

Short Answer

Expert verified
The magnitude of the linear impulse is \(364 \mathrm{~kg \cdot m/s}\) and the average total force experienced by the passenger is \(437.5 \mathrm{~N}\)

Step by step solution

01

Calculate Linear Impulse

Impulse can be calculated using the formula Impulse = mass * change in velocity. The change in velocity is the final velocity minus the initial velocity. Here, the final velocity of the car is \(5.20 \mathrm{~m/s}\) and the initial velocity is \(0 \mathrm{~m/s}\), so the change in velocity is \(5.20 \mathrm{~m/s}\). The mass of the passenger is \(70.0 \mathrm{~kg}\), so the impulse is \(70.0 \mathrm{~kg} * 5.20 \mathrm{~m/s} = 364 \mathrm{~kg \cdot m/s}\)
02

Calculate Acceleration

Acceleration is the change in velocity divided by the change in time. From the previous step we know that the change in velocity is \(5.20 \mathrm{~m/s}\). The time it takes for the car to accelerate is given as \(0.832 \mathrm{~s}\), so the acceleration is \(5.20 \mathrm{~m/s} / 0.832 \mathrm{~s} = 6.25 \mathrm{~m/s^2}\)
03

Calculate Average Total Force

Force is mass times acceleration. From the previous steps we know that the mass of the passenger is \(70.0 \mathrm{~kg}\) and the acceleration of the car is \(6.25 \mathrm{~m/s^2}\). Therefore, the force experienced by the passenger is \(70.0 \mathrm{~kg} * 6.25 \mathrm{~m/s^2} = 437.5 \mathrm{~N}\)

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Key Concepts

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

Impulse Formula
Understanding the impulse formula is crucial in physics, as it helps describe how the momentum of an object changes when it's subjected to a force over a period of time. Impulse is defined as the product of the average force applied to an object and the time duration over which the force is applied. This is mathematically represented as:

\[ \text{Impulse} = \text{Force} \times \text{Time} \]
or, since force is mass times acceleration, \[ \text{Impulse} = \text{Mass} \times \Delta\text{Velocity} \]
where \( \Delta\text{Velocity} \) is the change in velocity.

In the case of our exercise, where a passenger in a car experiences a change in velocity from rest to \(5.20 \text{ m/s}\), the impulse can be easily calculated by multiplying the mass of the passenger by this change in velocity. The formula captures the idea that if you push something for a longer time, or with more force, the total impulse delivered, and hence the object's change in momentum, will be greater. This understanding of impulse is essential in analyzing various real-world scenarios, such as car crashes, jumps in sports, or the firing of projectiles.
Acceleration Calculation
Acceleration represents how quickly an object speeds up, slows down, or changes direction. It is a vector quantity, meaning it has both magnitude and direction. The standard formula to calculate linear acceleration is:

\[ \text{Acceleration} = \frac{\Delta\text{Velocity}}{\Delta\text{Time}} \]
where \( \Delta\text{Velocity} \) is the change in velocity and \( \Delta\text{Time} \) is the time period over which this change occurs.

By applying this formula to our exercise, where a car accelerates over 0.832 seconds, we can find the acceleration by dividing the velocity change of \(5.20 \text{ m/s}\) by the time interval. Precise calculation of acceleration is vital as it is used in numerous physics equations and principles, including determining the forces acting on moving objects as seen in our application of the concept to find the average total force on a passenger.
Newton's Second Law
Newton's Second Law of Motion is a cornerstone of classical mechanics and states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This is commonly written as:

\[ \text{Force} = \text{Mass} \times \text{Acceleration} \]
According to this principle, if you know the mass of an object and its acceleration, you can calculate the force. In our exercise, applying Newton's Second Law helps us find the average total force exerted on the passenger as the car speeds up.

Understanding Newton's Second Law is fundamental to predicting how objects will move under the influence of various forces. It ties into the impulse formula as well, since force can also be understood as the change in momentum (impulse) per unit of time. Consequently, this law provides the framework for analyzing motion and is a critical concept in not only physics but also in engineering, aeronautics, and various other scientific and practical fields.

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Most popular questions from this chapter

A \(6.5 .0-\mathrm{kg}\) person throws a \(0.0450-\mathrm{kg}\) snowball forward with a ground speed of \(30.0 \mathrm{~m} / \mathrm{s} .\) A second person, with a mass of \(60.0 \mathrm{~kg}\), catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of \(2.50 \mathrm{~m} / \mathrm{s}\), and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

A billiard ball mowing at \(5.00 \mathrm{~m} / \mathrm{s}\) strikes a stationary ball of the same mass. After the collision, the first ball moves at \(4.39 \mathrm{~m} / \mathrm{s}\) at an angle \(0 \mathrm{f} 30^{\circ}\) with respect to the original line of motion. (a) Find the velocity (magnitude and direction) of the second ball after collision. (b) Was the collision inelastic or elastic?

Q|C Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of \(12 \mathrm{~m} / \mathrm{s}\), and the mass of rain per second striking the roof is \(0.035 \mathrm{~kg} / \mathrm{s}\). (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof. (b) If hailstones having the same mass as the raindrops fall on the roof at the same rate and with the same speed, how would the average force on the roof compare to that found in part (a)?

This is a symbolic version of Problem 23. A girl of mass \(m_{G}\) is standing on a plank of mass \(m_{p}\). Both are originally at rest on a frozen lake that constitutes a frictionless, flat surface. The girl begins to walk along the plank at a constant velocity \(x_{C P}\) to the right relative to the plank. (The subscript \(G P\) denotes the girl relative to plank.) (a) What is the velocity \(v_{P}\) of the plank relative to the surface of the ice? (b) What is the girl's velocity \(v_{C I}\) relative to the ice surface?

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20-kg puck has a speed of 1.0 m/s at an angle of u 5 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision. (b) Find the fraction of kinetic energy lost in the collision.

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