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17 PE

Expert-verifiedFound in: Page 82

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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 (\({\bf{9}}{\bf{.80}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) ) by taking its ratio to the acceleration of gravity.**

(a) The acceleration of the rocket sled is \({\bf{56}}{\bf{.4}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) or \({\bf{5}}{\bf{.8g}}\).

(b) The deceleration of the rocket sled is \({\bf{ - 201}}{\bf{.43}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) or \({\bf{ - 20}}{\bf{.6g}}\).

(a)

Given Data:

The initial speed of rocket is \(u = 0\;{\rm{m}}/{\rm{s}}\)

The final speed of rocket is \(v = 282\;{\rm{m}}/{\rm{s}}\)

The time for acceleration of rocket sled is \({t_a} = 5\;{\rm{s}}\)

The time for deceleration of rocket sled is \({t_d} = 1.40\;{\rm{s}}\)

**The acceleration of the rocket sled is variation in speed for increase in speed over time and deceleration of rocket sled is decrease in speed over time.**

The acceleration of rocket is given as

\(a = \frac{{v - u}}{{{t_a}}}\)

Here, \(a\) is the acceleration of rocket sled.

Substitute all the values in the above equation.

\(\begin{array}{l}a = \frac{{282\;{\rm{m}}/{\rm{s}} - 0\;{\rm{m}}/{\rm{s}}}}{{5\;{\rm{s}}}}\\a = 56.4\;{\rm{m}}/{{\rm{s}}^2}\\a = \left( {56.4\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {\frac{g}{{9.80\;{\rm{m}}/{{\rm{s}}^2}}}} \right)\\a = 5.8g\end{array}\)

Therefore, the acceleration of rocket sled is \(56.4\;{\rm{m}}/{{\rm{s}}^2}\) or \(5.8g\).

(b)

The deceleration of rocket is given as

\(f = \frac{{u - v}}{{{t_d}}}\)

Here, \(f\) is the deceleration of rocket sled.

Substitute all the values in the above equation.

\(\begin{array}{l}f = \frac{{0\;{\rm{m}}/{\rm{s}} - 282\;{\rm{m}}/{\rm{s}}}}{{1.40\;{\rm{s}}}}\\f = - 201.43\;{\rm{m}}/{{\rm{s}}^2}\\f = \left( { - 201.43\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {\frac{g}{{9.80\;{\rm{m}}/{{\rm{s}}^2}}}} \right)\\f = - 20.6g\end{array}\)

Therefore, the deceleration of rocket sled is \( - 201.43\;{\rm{m}}/{{\rm{s}}^2}\) or \( - 20.6g\).

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