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Chapter 2: Problem 25

During an auto accident, the vehicle's airbags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, airbags produce a maximum acceleration of \(60 g\) that lasts for only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g}\) ?

Short Answer

Expert verified
The person travels approximately 0.3813 meters while coming to a complete stop.

Step by step solution

01

Understanding the Problem

We are given the acceleration \(a = 60g\) and time \(t = 36 ms\) during which the acceleration acts. We want to find the distance travelled \(d\) during this time. The initial velocity of the driver is not specified, it is reasonable to assume that the airbag fully stops the driver, so the final velocity is 0. We can use the equation \( d = v_i * t + 0.5 * a * t^2\).
02

Convert g and ms to m/s^2 and seconds

Gravity \(g\) on Earth is \(9.8 m/s^2\), and \(1 ms = 1/1000 s\). So we substitute these values to get \(a = 60 * 9.8 m/s^2 = 588 m/s^2\) and \(t = 36/1000 s = 0.036 s\).
03

Substitute the values in the formula

Now we substitute the values \(a = 588 m/s^2\), \(t = 0.036 s\), and \(v_i = 0\) in the formula \(d = v_i * t + 0.5 * a * t^2\) to get \(d = 0*0.036 + 0.5 * 588 * (0.036)^2 = 0.381312 m\).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the causes of this motion. It focuses on the relationship between variables such as displacement, velocity, acceleration, and time. Understanding kinematics is crucial as it forms the foundation for solving physics problems involving motion.
In the given exercise, we are dealing with kinematic equations to analyze the motion of a person during a car accident. Specifically, we use the equation that relates distance, initial velocity, acceleration, and time:
  • \( d = v_i \times t + 0.5 \times a \times t^2 \)
This equation helps us understand how all the variables interact to determine the distance the person travels while coming to a stop.
Constant acceleration
In physics, constant acceleration refers to a situation where an object's velocity changes at a steady rate over time. This is an essential concept because it simplifies the analysis of motion and allows us to use straightforward equations.
In the context of this problem, the airbag provides a constant acceleration of \(60g\), which translates to \(588\) m/s虏. This means that for every second the airbag deploys, the person's velocity changes steadily by \(588\) meters per second. Understanding this constancy in acceleration is key to accurately calculating how far the person travels in the short time of airbag deployment.
Airbag deployment
Airbag deployment is a critical safety feature in automobiles, designed to protect passengers during a collision. The airbag inflates rapidly to cushion and slow down the passengers more gently than other parts of the vehicle like the windshield or steering wheel.
By deploying, the airbag provides the majority of the force needed to decelerate the passenger in a controlled manner. This reduces the likelihood of injury. In our exercise, we see how the airbag provides an acceleration of \(60g\) over a span of \(36\) milliseconds. This brief interval is enough to significantly reduce the speed of the passenger, bringing them to a stop over a controlled distance.
Distance calculation
The calculation of distance in this scenario requires us to apply the kinematic equation mentioned earlier. It's crucial to have all units in a consistent measure to correctly compute the result.
First, convert the given acceleration from \(g\) to meters per second squared (\(588\) m/s虏) and the time from milliseconds to seconds (\(0.036\) s). Then, substituting these values into the equation \(d = v_i \times t + 0.5 \times a \times t^2 \), where the initial velocity is assumed to be zero, yields a distance:
  • \(d = 0 \times 0.036 + 0.5 \times 588 \times 0.036^2 \approx 0.381312 \text{ meters}\)
This represents the distance traveled by the person while coming to a complete halt, highlighting the efficacy of airbag technology in enhancing passenger safety.

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Most popular questions from this chapter

An object is moving along the \(x\) -axis. At \(t=0\) it is at \(x=0 .\) Its \(x\) -component of velocity \(v_{x}\) as a function of time is given by \(v_{x}(t)=\alpha t-\beta t^{3},\) where \(\alpha=8.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(\beta=4.0 \mathrm{~m} / \mathrm{s}^{4}\) (a) At what nonzero time \(t\) is the object again at \(x=0 ?\) (b) At the time calculated in part (a), what are the velocity and acceleration of the object (magnitude and direction)?

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of \(960 \mathrm{~m}\) above the earth's surface. The rocket's engines give the rocket an upward acceleration of \(16.0 \mathrm{~m} / \mathrm{s}^{2}\) during the time \(T\) that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. What must be the value of \(T\) in order for the rocket to reach the required altitude?

A 15 kg rock is dropped from rest on the earth and reaches the ground in \(1.75 \mathrm{~s}\). When it is dropped from the same height on Saturn's satellite Enceladus, the rock reaches the ground in \(18.6 \mathrm{~s}\). What is the acceleration due to gravity on Enceladus?

The acceleration of a bus is given by \(a_{x}(t)=\alpha t,\) where \(\alpha=1.2 \mathrm{~m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{~s}\) is \(5.0 \mathrm{~m} / \mathrm{s}\) what is its velocity at time \(t=2.0 \mathrm{~s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{~s}\) is \(6.0 \mathrm{~m},\) what is its position at time \(t=2.0 \mathrm{~s} ?(\mathrm{c})\) Sketch \(a_{y}-t\) \(v_{y}-t,\) and \(x-t\) graphs for the motion.

A rock moving in the \(+x\) -direction with speed \(16.0 \mathrm{~m} / \mathrm{s}\) has a net force applied to it at time \(t=0,\) and this produces a constant acceleration in the \(-x\) -direction that has magnitude \(4.00 \mathrm{~m} / \mathrm{s}^{2}\). For what three times \(t\) after the force is applied is the rock a distance of \(24.0 \mathrm{~m}\) from its position at \(t=0 ?\) For each of these three values of \(t,\) what is the velocity (magnitude and direction) of the rock?
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