Question

A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 g’s. Calculate the force on a 65-kg person accelerating at this rate. What distance is traveled if brought to rest at this rate from$95\mathrm{km}/\mathrm{h}$.

Short answer

The force acting on the person is $\left(19129.5\mathrm{N}\right)$.

The distance traveled by the automobile is $1.18\mathrm{m}$.

Step-by-step solution

Step 1. Understanding the application of Newton’s law on an automobile 

The force exerted on the person can be calculated with the help of Newton’s second law.

The product of the acceleration of the automobile and the mass of the person gives the force.

The distance traveled by the automobile can be estimated with the help of Newton’s third law.

Step 2. Identification of given data 

The given data can be listed below as:

• The mass of the person is $m=65\mathrm{kg}$.
• The deceleration of the automobile is $a=30g$.
• The acceleration due to gravity is $g=9.81\mathrm{m}/{\mathrm{s}}^2$.
• The initial velocity of the automobile is

$v_i=95\mathrm{km}/\mathrm{h}\times \frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{km}/\mathrm{h}}=26.39\mathrm{m}/\mathrm{s}$.

Step 3. Determination of the force exerted on the person

From Newton’s second law, the force exerted on the person can be given as:

$F=ma$

Here, m is the mass of the person, and a is the acceleration of the automobile. The acceleration of the automobile is taken to be positive because the person is accelerating at $30g$.

Substitute the values as 65 kg for m, $\left(30g\right)$for a, and $9.81\mathrm{m}/{\mathrm{s}}^2$for g in the above equation.

$\begin{array}{c}F=65\mathrm{kg}\times 30g\\ =65\mathrm{kg}\times 30\times 9.81\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =19129.5\mathrm{N}\end{array}$

Thus, the force acting on the person is $\left(19129.5\mathrm{N}\right)$.

Step 4. Determination of the distance traveled by the automobile

From Newton’s third law, the final velocity of the automobile can be expressed as:

$\begin{array}{c}{v}_f^2=v_i^2+2ad\\ d=\frac{v_f^2-v_i^2}{2a}\end{array}$

Here, $v_f$is the final velocity of the automobile. The final velocity of the automobile becomes 0 m/s as the automobile comes to rest. The acceleration is taken to be negative because the automobile is decelerating at $30g$.

Substitute the values as $0\mathrm{m}/\mathrm{s}$ for $v_f$, $26.39\mathrm{m}/\mathrm{s}$ for $v_i$, $\left(-30g\right)$ for a, and $9.81\mathrm{m}/{\mathrm{s}}^2$for g in the above equation.

$\begin{array}{c}d=\frac{{\left(0\mathrm{m}/\mathrm{s}\right)}^2-{\left(26.39\mathrm{m}/\mathrm{s}\right)}^2}{2\times \left(-30g\right)}\\ =\frac{{\left(0\mathrm{m}/\mathrm{s}\right)}^2-{\left(26.39\mathrm{m}/\mathrm{s}\right)}^2}{2\times \left(-30\right)\times 9.81\mathrm{m}/{\mathrm{s}}^2}\\ =1.18\mathrm{m}\end{array}$

Thus, the distance traveled by the automobile is $1.18\mathrm{m}$.