Chapter 2, Problem 57GP

(a)

To determine

The average deceleration of person when she is jumping into firefighter’s safety net.

Answer to Problem 57GP

Solution: $a=-176.3\text{4}\text{m}/{\text{s}}^{\text{2}}$

Explanation of Solution

Given:

Initial velocity $\left(u\right)=0\text{m}/\text{s}$

Height $\left(h\right)=18\text{m}$

Acceleration $\left(a\right)=9.8\text{m}/{\text{s}}^{\text{2}}$

Formula Used:

Third equation of motion:

  $v^2-u^2=2as$

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

Calculations:

Here, $s=h$

  $\begin{array}{l}{v}^2=u^2+2gh\\ v^2=0+2\times 9.8\times 18\end{array}$

  $v=\sqrt{2\times 9.8\times 18}$

  $v=18.78\text{m}/\text{s}$

The average deceleration of person after making contact with the safety net.

Before making contact with safety net, person velocity is $v=18.78\text{m}/\text{s}$

Final velocity in last section will be considered as initial velocity of person in this section

Initial velocity $\left(u\text{'}\right)=18.78\text{m}/\text{s}$

Final velocity $\left(v\right)=0\text{m}/\text{s}$

Distance $\left(s\right)=1\text{m}$

  $v^2-u^2=2as$

  $a=\frac{-u^2}{2s}$

  $a=\frac{-{\left(18.78\right)}^2}{2\times 1}$

  $a=-176.34\text{m}/{\text{s}}^{\text{2}}$

Conclusion:

The average deceleration experienced by her is $a=-176.34\text{m}/{\text{s}}^{\text{2}}$ .

(b)

To determine

Whether the net should be tightened or loosened for smaller deceleration.

Answer to Problem 57GP

Solution:The net should be loosened.

Explanation of Solution

From the third equation of motion:

  $a=\frac{-u^2}{2s}\left(\because v=0\right)$

If net is tightened, the value of displacement (s) will decrease which will increase the value of deceleration. Thus, it is saferto loosenthe net so thatthe person has greater displacement, decelerates slowly and hence can be brought to rest more safely.