Ch 02: Motion Along a Straight Line

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 02: Motion Along a Straight Line

Problem 23

Chapter 2, Problem 23

The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than $250$ m/s². If you are in an automobile accident with an initial speed of $105$ km/h ( $65$ mi/h) and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

Verified step by step guidance

First, convert the initial speed from kilometers per hour to meters per second. Use the conversion factor: 1 km/h = 0.27778 m/s. Therefore, 105 km/h is equivalent to 105 * 0.27778 m/s.

Next, identify the final speed after the stop, which is 0 m/s since the car comes to a complete stop.

Use the formula for acceleration: $ a = \frac{v^2 - u^2}{2s} $, where $ v $ is the final velocity, $ u $ is the initial velocity, $ a $ is the acceleration, and $ s $ is the stopping distance.

Rearrange the formula to solve for the stopping distance $ s $: $ s = \frac{v^2 - u^2}{2a} $. Substitute $ v = 0 $, $ u $ as the converted initial speed, and $ a = 250 $ m/s².

Calculate the stopping distance $ s $ using the values substituted into the rearranged formula. This will give you the minimum distance over which the airbag must stop you to ensure survival.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Acceleration

Acceleration is the rate of change of velocity of an object. It is a vector quantity, having both magnitude and direction, and is measured in meters per second squared (m/s²). In the context of the question, understanding acceleration is crucial to determine the deceleration needed to safely stop a person during a crash.

Kinematics Equations

Kinematics equations describe the motion of objects without considering the forces that cause the motion. They relate displacement, initial velocity, final velocity, acceleration, and time. For this problem, the equation v² = u² + 2as can be used to find the stopping distance, where v is final velocity, u is initial velocity, a is acceleration, and s is displacement.

Conversion of Units

Conversion of units is essential for ensuring consistency in calculations. In this problem, the initial speed is given in kilometers per hour, which must be converted to meters per second to match the units of acceleration. This involves multiplying the speed by a conversion factor of 1000/3600 to convert km/h to m/s.

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Unit Conversions

Related Practice

Textbook Question

An antelope moving with constant acceleration covers the distance between two points $70.0$ m apart in $6.00$ s. Its speed as it passes the second point is $15.0$ m/s. What is its acceleration?

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Textbook Question

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then 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 $20$ m/s ( $45$ mi/h) when it reaches the end of the $120$ -m-long ramp. How much time does it take the car to travel the length of the ramp?

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Textbook Question

In the fastest measured tennis serve, the ball left the racquet at $73.14$ m/s. A served tennis ball is typically in contact with the racquet for $30.0$ ms and starts from rest. Assume constant acceleration. What was the ball's acceleration during this serve?

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Textbook Question

The fastest measured pitched baseball left the pitcher's hand at a speed of $45.0$ m/s. If the pitcher was in contact with the ball over a distance of $1.50$ m and produced constant acceleration, what acceleration did he give the ball?

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