Chapter 4, Problem 15P

Section (I)

To determine

The acceleration of a car that must be stopped on a dime.

Answer to Problem 15P

Solution:

The acceleration of the car is -2807.4 m/s²

Explanation of Solution

Given:

Initial velocity of the car, $V_i=35\text{km/h}$

Final velocity of the car, $V_f=0\text{m/s}$

Diameter of the dime, $d=1.7\text{cm}$

Formula:

Third equation of motion is:

  $v^2=u^2+2as$

Where, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

Calculation:

Convert unit of initial velocity from km/h to m/s.

  $\begin{array}{l}{V}_i=35\text{km/h}\\ {\text{V}}_i=35\times \frac{1000\text{m}}{3600\text{s}}\text{=9}\text{.77}\text{m/s}\end{array}$

Convert diameter of dime from cm to m.

  $\begin{array}{l}d=1.7\text{cm}\\ d=1.7\times {10}^{-2}\text{m}\\ d\text{ = }0.017\text{ m}\end{array}$

Substitute the given values in the equation of motion and solve for acceleration (a).

  $\begin{array}{l}{V}_f^2=V_i^2+2ad\\ \begin{array}{l}{0}^2={\left(9.77\right)}^2+2a\left(0.017\right)\hfill \\ a=-2807.4{\text{ m/s}}^{\text{2}}\hfill \end{array}\end{array}$

Conclusion: When the car stops on thedime, its acceleration is 2807.4 m/s² in the backward direction.

Section(II)

To determine

The number of g’s in the value of acceleration calculated.

Answer to Problem 15P

Solution:

Acceleration is approximately 286 times the acceleration due to gravity.

Explanation of Solution

Given:

Acceleration of the car, $a=-2807.4{\text{ m/s}}^{\text{2}}$

Formula Used:

Number of g’s = $\frac{a}{g}$

Where, a is the acceleration and g is the gravitational acceleration.

Calculation:

Substitute the values and solve.

Number of g’s

  $\begin{array}{l}=\frac{a}{g}\\ =\frac{-2807.4}{-9.8}\\ =286.17\end{array}$

Conclusion: The acceleration is approximately 286 times the acceleration due to gravity.

Section (III)

To determine

The force experienced by an occupant of the car in stopping the car on a dime.

Answer to Problem 15P

Solution:

The force experienced by the occupant is 190903 N

Explanation of Solution

Given:

Mass of the occupant, $m=68\text{ kg}$

Acceleration of the car, $a=-2807.4{\text{m/s}}^{\text{2}}$

Formula:

From Newton’s second law of motion:

  $F=ma$

Here, m is the mass and a is the acceleration.

Calculation:

Substitute the values and solve for force ( F ).

  $\begin{array}{l} F=\left(68\right)\left(-2807.4\right)\hfill \\ F=190903\text{ N}\hfill \end{array}$

Conclusion: Due to this acceleration occupant experiences a force of 190903 N.