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When a car is hit from behind, its passengers undergo sudden forward acceleration, which can cause a severe neck injury known as whiplash. During normal acceleration, the neck muscles play a large role in accelerating the head so that the bones are not injured. But during a very sudden acceleration, the muscles do not react immediately because they are flexible; most of the accelerating force is provided by the neck bones. Experiments have shown that these bones will fracture if they absorb more than 8.0 J of energy. (a) If a car waiting at a stoplight is rear-ended in a collision that lasts for 10.0 m/ s, what is the greatest speed this car and its driver can reach without breaking neck bones if the driver’s head has a mass of 5.0 kg (which is about right for a 70-kg person)? Express your answer in m/s and in mi/h. (b) What is the acceleration of the passengers during the collision in part (a), and how large a force is acting to accelerate their heads? Express the acceleration in $\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$and in g’s.

Step by step solution

01

Given data

Energy = 8.0 J

Collision = $10.0\mathrm{m}.{\mathrm{s}}^2$

02

Solution

Use the law of conservation energy to calculate the greatest speed of the car.

According to the law of conservation of energy,

${\mathrm{E}}_{\mathrm{i}}={\mathrm{E}}_{\mathrm{f}}$

Here, ${\mathrm{E}}_{\mathrm{i}}$is the initial energy and ${\mathrm{E}}_{\mathrm{f}}$is the final energy.

03

(a) Find the greatest speed this car

According to the law of conservation of energy,

$\frac{1}{2}{\mathrm{mv}}_{\max }^2=8.0J$

Rearrange the above equation for speed and substitute 5.0 kg for min above equation as follows:

$$\begin{gathered} {\mathrm{v}}_{\max }=\sqrt{\frac{2×\left(8.0\mathrm{J}\right)}{5.0\mathrm{kg}}} \\ =1.8\mathrm{m}/\mathrm{s}\left(\frac{2.23694\mathrm{mi}/\mathrm{h}}{1\mathrm{m}/\mathrm{s}}\right) \\ =4.0\mathrm{mi}/\mathrm{h} \end{gathered}$$

Therefore, the greatest speed is 1.8 m/s or 4.0 mi/h.

04

(b) Find the acceleration of the car

According to the kinematic equation, the acceleration of the car is,

$\mathrm{a}=\frac{{\mathrm{v}}_{\max }-\mathrm{u}}{\mathrm{t}}$

Substitute 0m/s for u , 4.0 mi/h for $v_{max}$, and 10 ms for t in the above equation as follows:

$$\begin{gathered} \mathrm{a}=\frac{\left(4.0\mathrm{mi}/\mathrm{h}\right)\left({\displaystyle \frac{1\mathrm{m}/\mathrm{s}}{2.23694\mathrm{mi}/\mathrm{h}}}\right)}{10\mathrm{ms}\left({\displaystyle \frac{{10}^{-3}\mathrm{s}}{1\mathrm{ms}}}\right)} \\ =180\mathrm{m}.{\mathrm{s}}^{-2} \end{gathered}$$

Therefore, the acceleration of the car is $180\mathrm{m}.{\mathrm{s}}^{-2}$.

The acceleration can be written as follows:

$$\begin{gathered} \mathrm{a}=180\mathrm{m}.{\mathrm{s}}^{-2}\left(\frac{\mathrm{g}}{9.8\mathrm{m}.{\mathrm{s}}^{-2}}\right) \\ =18.4\mathrm{g} \end{gathered}$$

Therefore, the acceleration in terms of is 18.4 g.

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