Question

: Most important in an investigation of an airplane crash by the U.S. National Transportation Safety Board is the data stored on the airplane’s flight-data recorder, commonly called the “black box” in spite of its orange coloring and reflective tape. The recorder is engineered to withstand a crash with an average deceleration of magnitude $\mathbf{3400}\mathbf{}\mathbf{g}$during a time interval of $\mathbf{6}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{ms}$. In such a crash, if the recorder and airplane have zero speed at the end of that time interval, what is their speed at the beginning of the interval?

Short answer

The the speed of the recorder and the airplane at the beginning of the interval is$216.6\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given data

The average deceleration,$a=-3400\mathrm{g}$

Time interval,role="math" localid="1654752848004" $t=6.50\mathrm{ms}$

Understanding the kinematic equations

Kinematic equations describe the motion of an object with constant acceleration. These equations can be used to determine the acceleration, velocity or distance.

The expression for the kinematic equations of motion are given as follows:

$\mathrm{v}={\mathrm{v}}_0+\mathrm{at}$

… (i)

Here,${\mathrm{v}}_0$is the initial velocity, is the final velocity, role="math" localid="1654752999268" $t$is the time, $\mathrm{a}$is the acceleration.

Determination of the speed at the beginning of the interval

The final speed of the airplane is zero, that is$\mathrm{v}=0$

Using equation (i) the initial speed can be calculated as follows:

role="math" localid="1654753374414" $$\begin{gathered} v_0=\mathrm{v}-\mathrm{at} \\ =0-\left(-3400\mathrm{g}\right)\times \left(6.50ms\times \frac{1\mathrm{s}}{100\mathrm{ms}}\right) \\ =\left(3400\times 9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(6.50\times {10}^3\mathrm{s}\right) \\ =216.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the recorder and the airplane at the beginning of the interval is$216.6\mathrm{m}/\mathrm{s}$.