Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his
(a) acceleration and
(b) deceleration.
Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.
Solution:
Part A
The formula for acceleration is
\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & = \frac{v_f-v_0}{t_f-t_0} \\
\end{align*}Substituting the given values
\begin{align*}
\overline{a} & =\frac{282\:\text{m/s}-0\:\text{m/s}}{5.00\:\sec } \\
\overline{a} & =56.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
The deceleration is
\begin{align*}
\overline{a} & =\frac{0\:\text{m/s}-282\:\text{m/s}}{1.40\:\text{s}} \\
\overline{a} & =-201.43\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}In expressing the computed values in terms of g, we just divide them by 9.80.
The acceleration is
\overline{a}=\frac{56.4}{9.80}=5.76\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)The deceleration is
\overline{a}=\frac{201.43}{9.80}=20.55\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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