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Chapter 2: Problem 32

A world's land speed record was set by Colonel John P. Stapp when in March 1954 he rode a rocket-propelled sled that moved along a track at \(1020 \mathrm{~km} / \mathrm{h} .\) He and the sled were brought to a stop in \(1.4 \mathrm{~s}\). (See Fig. 2-7.) In terms of \(g\), what acceleration did he experience while stopping?

Short Answer

Expert verified
Colonel Stapp experienced an acceleration of approximately \(-20.63g\).

Step by step solution

01

Convert Speed to Meters Per Second

First, transform the speed from kilometers per hour to meters per second. We know that 1 km/h is equal to 1/3.6 m/s. Therefore, for a speed of 1020 km/h, we have:\[ 1020 \times \frac{1}{3.6} = 283.33 \text{ m/s} \]
02

Calculate Deceleration

Next, calculate the deceleration needed to stop the sled. The formula for acceleration (or deceleration) is \( a = \frac{v_f - v_i}{t} \), where \(v_f\) is the final velocity (0 m/s since the sled stops), \(v_i\) is the initial velocity (283.33 m/s), and \(t\) is the time taken to stop (1.4 s). Thus:\[ a = \frac{0 - 283.33}{1.4} = -202.38 \text{ m/s}^2 \]
03

Express Deceleration in Terms of g

Finally, convert the deceleration from m/s虏 to the unit \(g\), where \( g \approx 9.81 \text{ m/s}^2 \). Therefore, the acceleration in terms of \(g\) is:\[ \frac{-202.38}{9.81} \approx -20.63g \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Decimals Conversion
Understanding how to convert decimals is crucial when dealing with physics problems. Decimal conversion often comes up when translating speeds, distances, and other measurements from one unit to another. Take for instance the conversion of kilometers per hour (km/h) into meters per second (m/s), a common task in physics calculations. Here's how it's done:
  • Recognize the basic conversion factor: 1 km/h equals approximately 1/3.6 m/s.
  • To convert a speed of 1020 km/h into m/s, multiply 1020 by 1/3.6.
  • This gives us 283.33 m/s, making it ready for further computations like acceleration.
These steps are essential for making your physics calculations consistent and accurate.
Deceleration Calculation
Deceleration refers to the reduction in speed or velocity, a key concept when analyzing movement. In physics, deceleration can be calculated using the formula for acceleration: \[ a = \frac{v_f - v_i}{t} \] where:
  • \(v_f\) is the final velocity.
  • \(v_i\) is the initial velocity.
  • \(t\) is the time over which the change occurs.
For all practical purposes, deceleration is regarded as a negative acceleration, implying an object is slowing down. In the exercise example:
  • The initial velocity, \(v_i\), was 283.33 m/s, and the sled came to a stop (\(v_f = 0\) m/s).
  • The time required to stop was 1.4 seconds.
  • Plug these values into the formula to find \( a = \frac{0 - 283.33}{1.4} = -202.38 \text{ m/s}^2 \), indicating rapid deceleration.
Understanding deceleration helps to comprehend how forces act to stop moving objects.
Acceleration Unit Conversion
When you calculate acceleration, sometimes you need to express it in different units, such as converting meters per second squared (m/s虏) to \(g\) units. \(g\) represents gravitational acceleration, where 1 \(g\) is approximately equal to 9.81 m/s虏. To convert acceleration to \(g\)-units:
  • Use the formula: \( a_g = \frac{a}{ g} \), where \(a\) is acceleration in m/s虏.
  • From the example, \(-202.38 \text{ m/s}^2\) is divided by 9.81 \(\text{m/s}^2\).
  • This results in approximately \(-20.63g\).
Expressing acceleration in terms of \(g\) makes it relatable to everyday experiences of gravity, providing a more intuitive grasp of the forces involved.

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Most popular questions from this chapter

You are arguing over a cell phone while trailing an unmarked police car by \(25 \mathrm{~m}\); both your car and the police car are traveling at \(110 \mathrm{~km} / \mathrm{h}\). Your argument diverts your attention from the police car for \(2.0 \mathrm{~s}\) (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that \(2.0 \mathrm{~s}\), the police officer begins braking suddenly at \(5.0 \mathrm{~m} / \mathrm{s}^{2} .(\mathrm{a})\) What is the separation between the two cars when your attention finally returns? Suppose that you take another \(0.40 \mathrm{~s}\) to realize your danger and begin braking. (b) If you too brake at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\), what is your speed when you hit the police car?

An automobile travels on a straight road for \(40 \mathrm{~km}\) at \(30 \mathrm{~km} / \mathrm{h}\). It then continues in the same direction for an- other \(40 \mathrm{~km}\) at \(60 \mathrm{~km} / \mathrm{h}\). (a) What is the average velocity of the car during the full \(80 \mathrm{~km}\) trip? (Assume that it moves in the positive \(x\) direction.) (b) What is the average speed? (c) Graph \(x\) versus \(t\) and indicate how the average velocity is found on the graph.

A basketball player grabbing a rebound jumps \(76.0 \mathrm{~cm}\) vertically. How much total time (ascent and descent) does the player spend (a) in the top \(15.0 \mathrm{~cm}\) of this jump and (b) in the bottom \(15.0 \mathrm{~cm} ?\) (The player seems to hang in the air at the top.)

A car moving with constant acceleration covered the distance between two points \(60.0 \mathrm{~m}\) apart in \(6.00 \mathrm{~s}\). Its speed as it passed the second point was \(15.0 \mathrm{~m} / \mathrm{s}\). (a) What was the speed at the first point? (b) What was the magnitude of the acceleration? (c) At what prior distance from the first point was the car at rest? (d) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the car, from rest \((t=0)\).

A train started from rest and moved with constant acceleration. At one time it was traveling \(30 \mathrm{~m} / \mathrm{s}\), and \(160 \mathrm{~m}\) farther on it was traveling \(50 \mathrm{~m} / \mathrm{s}\). Calculate (a) the acceleration, (b) the time required to travel the \(160 \mathrm{~m}\) mentioned, \((\mathrm{c})\) the time required to attain the speed of \(30 \mathrm{~m} / \mathrm{s}\), and \((\mathrm{d})\) the distance moved from rest to the time the train had a speed of \(30 \mathrm{~m} / \mathrm{s}\). (e) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the train, from rest.
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