Chapter 3, Problem 97A

(a)

To determine

The description of the motion of the G’s fist during the first 10 ms.

Answer to Problem 97A

The motion of the G is decelerating and becomes zero at the time 10 ms.

Explanation of Solution

Introduction:

The slope of the position time graph describes the velocity and slope of the velocity time graph describes the acceleration of the G’s fist.

The position time graph shows that the G starts from the 7 cm initially and after hitting the wooden board it comes at rest near about in 5 ms. The velocity time graph shows that G starting with decelerating motion and stops accelerating near about 8 ms. The G again starts decelerating between the 8 ms to 10 ms.

Conclusion:

Thus, the G starts initially with some distance and decelerating between the 8 ms to 10 ms.

(b)

To determine

The slope of the velocity time graph to find the acceleration of the fist.

Answer to Problem 97A

The acceleration of the fist is $2600\text{m}/{\text{s}}^2$ .

Explanation of Solution

Given Info:

The initial velocity of the fist at $t_1=3\text{ms}$ is, $v_1=-13\text{}\text{m}/\text{s}$ .

The final velocity of the fist at $t_2=8\text{ms}$ is, $v_2=0\text{m}/\text{s}$

Formula Used:

The expression to calculate the slope of the velocity time graph is,

$a=\frac{v_2-v_1}{t_2-t_1}$

Calculation:

Substitute all the values in the above expression.

$\begin{array}{c}a=\frac{0\text{m}/\text{s}-\left(-13\text{m}/\text{s}\right)}{8\text{ms}-3\text{ms}}\\ =\frac{13\text{m}/\text{s}}{\left(5\text{ms}\times \frac{1\text{s}}{1000\text{ms}}\right)}\\ =2600\text{m}/{\text{s}}^2\end{array}$

Thus, the acceleration of the fist is $2600\text{m}/{\text{s}}^2$ .

(c)

To determine

The acceleration of the fist as a multiple of the free fall acceleration.

Answer to Problem 97A

The acceleration of the fist as a multiple of the free fall acceleration is $265.31g$ .

Explanation of Solution

Given Info:

The magnitude of the free fall acceleration is, $g=9.8\text{m}/{\text{s}}^2$ .

Calculation:

The acceleration of the fist as the multiple of the free fall acceleration is,

$\begin{array}{c}a=2600\text{m}/{\text{s}}^2\times \frac{1g}{9.8\text{m}/{\text{s}}^2}\\ =265.31g\end{array}$

(d)

To determine

The area under the curve of the velocity time graph to find the displacement of the fist in first 6 ms.

Answer to Problem 97A

The displacement of the fist from the velocity time graph is $0.0645\text{m}$ that is $0.0005\text{m}$ less than the displacement reading from the position time graph.

Explanation of Solution

Introduction:

The area under the curve in the velocity-time graph describes the displacement of the object. The curve for the first three seconds is trapezoid and for last three seconds the curve becomes rectangle.

The area under the curve of the velocity time graph to find the displacement of the fist in first 6 ms is given as,

$\begin{array}{c}d=\frac{1}{2}\left(13\text{m}/\text{s}+4\text{m}/\text{s}\right)\left(3\text{ms}\times \frac{1\text{s}}{1000\text{ms}}\right)+\left(13\text{m}/\text{s}\right)\left(3\text{ms}\times \frac{1\text{s}}{1000\text{ms}}\right)\\ =0.0645\text{m}\end{array}$

The displacement of the fist from the position time graph is at $0.065\text{m}$ and the displacement of the fist from the velocity time graph is at $0.0645\text{m}$ . There is the difference of $0.0005\text{m}$ in the displacement reading between the position time graph and velocity time graph.

Conclusion:

The displacement of the fist from the velocity time graph is $0.0645\text{m}$ that is $0.0005\text{m}$ less than the displacement reading from the position time graph.