Chapter 4, Problem 6P

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 km/h?

To determine

The force on a person and the distance he traveled if brought to rest.

Answer to Problem 6P

Solution:

The force acting on a person is $1.9\times {10}^4\text{ N}$ and the distance traveled by him is $\text{1}\text{.2 m}$

Explanation of Solution

The average force on a person is calculated by using Newton’s second law of motion as given below:

$F=ma$………… (1)

Here, $m$ is the mass of the person and $a$ is its deceleration.

Now, by using Newton’s equation of motion the distance traveled by the person is given as follows:

$v^2-u^2=2as$………… (2)

Here, $s$ is the distance traveled by the person, $u$ is the final speed of the person and $v$ is its initial speed.

Given:

Mass of the person $m=65\text{ kg}$

Deacceleration $a=-30\text{ g}$

The initial speed of the person $u=95\text{ km/h}$

Final speed of the person $v=0\text{ km/h}$

Formula used:

$F=ma$

$v^2-u^2=2as$

Calculation:

Substitute, $m=65\text{ kg}$ and $a=-30\text{ g}$ in equation (1)so the force exerted by a person is given as below:

F=ma

  =(65 kg)(-30 g)

  =(65 kg)(-30)(9.8 m/s²)

  =1.9×10⁴ N

Now, we have to substitute, $u=95\text{ km/h = 26}\text{.38 m/s}$, $v=0\text{ km/h}$$a=-30\text{ g}$ in equation (2)so the distance traveled by a person is given as follows:

               v²-u²=2as

0-(26.38 m/s)²=2(-30×9.8)s

                      s=1.18 m

                       =1.2 m