Chapter 4, Problem 67GP

To determine

The distance travelled by the person before coming to rest and force on the man.

Answer to Problem 67GP

The force on the person is $20580\text{N}$ and the distance travelled by the person before coming to rest is $1.40\text{}\text{m}$ .

Explanation of Solution

Given info:

The deacceleration of the automobile is, $a=30g=30\left(9.8\text{m}/{\text{s}}^2\right)=294\text{m}/{\text{s}}^2$ .

The speed of the person is, $u=100\text{km}/\text{h}=\left(100\text{km}/\text{h}\right)\left(\frac{0.287\text{m}/\text{s}}{1\text{km}/\text{h}}\right)=28.7\text{m}/\text{s}$ .

The mass of the person is, $m=70\text{kg}$

Formula Used:

The expression to calculate the force on the person is,

  $F=ma$

The expression to calculate thedistance travelled by the person before coming to rest is,

  $d=\frac{u^2}{2a}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}F=\left(70\text{kg}\right)\left(294\text{m}/{\text{s}}^2\right)\\ =20580\text{N}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}d=\frac{{\left(28.7\text{m}/\text{s}\right)}^2}{2\left(294\text{m}/{\text{s}}^2\right)}\\ =1.40\text{}\text{m}\end{array}$

Conclusion:

Thus, the force on the person is $20580\text{N}$ and the distance travelled by the person before coming to rest is $1.40\text{}\text{m}$ .