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Chapter 2: Problem 45

A world's land speed record was set by Colonel John P. Stapp when, on March 19,1954, he rode a rocket-propelled sled that moved down a track at \(1020 \mathrm{~km} / \mathrm{h}\). He and the sled were brought to a stop in \(1.4 \mathrm{~s}\); see Fig. \(2-28 .\) What acceleration did he experience? Express your answer in terms of \(g(=9.8\) \(\mathrm{m} / \mathrm{s}^{2}\) ), the acceleration due to gravity. (Note that his body acts as an accelerometer, not a speedometer.)

Short Answer

Expert verified
The acceleration Colonel John P. Stapp experienced during his land speed record was approximately -20.67 g.

Step by step solution

01

Convert speed from km/h to m/s

To perform calculations accurately, convert the speed from kilometers per hour to meters per second. Since there are 1000 meters in a kilometer and 3600 seconds in an hour, this can be done by multiplying the speed by \(\frac{1000}{3600} = 0.2778\). Therefore, Colonel Stapp's speed is \(1020\times 0.2778= 283.6 m/s\).
02

Apply the formula for acceleration

The formula for acceleration is given by \(a=\frac{Δv}{Δt}\), where \(Δv\) is the change in velocity and \(Δt\) is the change in time. Since Colonel Stapp was brought to a complete stop, the final velocity would be 0 m/s. Considering his initial velocity was 283.6 m/s and the time period was 1.4 seconds, the acceleration can be calculated as follows: \(a= -\frac{283.6}{1.4}= -202.57 m/s^2\). The negative sign denotes that this is deceleration, or slowing down.
03

Express acceleration in terms of g

To express the acceleration in terms of g, the value for acceleration due to gravity which is 9.8 \(m/s^2\), we need to divide the obtained acceleration by g. Therefore, \(a= -\frac{202.57}{9.8}= -20.67 g\) approximately, meaning the deceleration experienced was about 20.67 times the acceleration due to gravity on earth.

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