17 PE Dr. John Paul Stapp was U.S. Air... [FREE SOLUTION]

Chapter 2: 17 PE (page 82)

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.

Short Answer

Expert verified

(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 \({\bf{ - 20}}{\bf{.6g}}\).

Step by step solution

Determination of acceleration of rocket sled

(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 \(5.8g\).

Determination of deceleration of rocket sled

(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 \( - 20.6g\).