/*! 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 17 A car leaves the road traveling ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 17

A car leaves the road traveling at \(110 \mathrm{km} / \mathrm{h}\) and hits a tree, coming to a stop in 0.14 s. What average force does a seatbelt exert on a \(60-\mathrm{kg}\) passenger during this collision?

Short Answer

Expert verified
The seatbelt exerts an average force of approximately \( -13097.4 \mathrm{N} \) on the passenger during the collision.

Step by step solution

01

Calculate the initial velocity

The car is moving at \(110 \mathrm{km} / \mathrm{h}\). However, this must be converted into metres per second (m/s) because the standard unit of velocity in physics is m/s. To convert km/h into m/s, multiply the value by \(1000/3600\). So, the initial velocity \(v_i\) is \(110 \times 1000 / 3600 = 30.56 \mathrm{m/s}\) .
02

Calculate the final velocity

As the car comes to a stop after hitting the tree, the final velocity \(v_f\) is \(0 \mathrm{m/s}\).
03

Determine the change in time

The time taken for the car to stop, or change in time \(\Delta t\), is given as \(0.14 \mathrm{s}\).
04

Calculate Acceleration

Acceleration \(a\) is the change in velocity divided by the time it takes for this change. In this case, acceleration is the final velocity subtracted from the initial velocity, divided by the time. Mathematically, this is \(a = (v_f - v_i) / \Delta t\). Substituting the values, we get \(a = (0 - 30.56) / 0.14 = -218.29 \mathrm{m/s^2}\). The negative sign indicates that this is a deceleration.
05

Determine Force

Finally, the average force \(F\) exerted on the passenger by the seatbelt can be determined by using the formula \(F = m \cdot a\), where \(m\) is the mass of the passenger and \(a\) is the acceleration. Thus, the force is \(60 \mathrm{kg} \times -218.29 \mathrm{m/s^2} = -13097.4 \mathrm{N}\). The negative sign indicates that this force is exerted in the opposite direction of the motion of the car.

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