Problem 32 - Fundamentals of Physics

Question

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?

Step-by-Step Solution

Answer

The acceleration was approximately 20.63 g.

Step 1: Convert Speed to Meters per Second

The first step is to convert the speed from kilometers per hour to meters per second because standard units for acceleration are in meters and seconds. Given: - The speed is 1020 km/h.To convert km/h to m/s, use the conversion factor: 1 km/h = \(\frac{1}{3.6}\) m/s.Thus, \[1020 \text{ km/h} \times \frac{1}{3.6} = 283.33 \text{ m/s}\]

Step 2: Calculate Deceleration

Now, calculate the deceleration (or negative acceleration) using the formula for acceleration:\[a = \frac{\Delta v}{\Delta t}\]Where:- \(\Delta v\) is the change in velocity.- \(\Delta t\) is the change in time.Given:- \(\Delta v = 283.33 - 0 = 283.33\) m/s (since the final velocity is 0 when stopped)- \(\Delta t = 1.4\) sThus,\[a = \frac{283.33}{1.4} = 202.38 \text{ m/s}^2\]

Step 3: Express Acceleration in Terms of g

To express the deceleration in terms of \( g \), where \( g \approx 9.81 \text{ m/s}^2 \) is the acceleration due to gravity:\[\text{Acceleration in terms of } g = \frac{202.38 \text{ m/s}^2}{9.81 \text{ m/s}^2}\]Calculating gives:\[\frac{202.38}{9.81} \approx 20.63\, g\]

Step 4: Conclusion

Thus, the acceleration experienced by Colonel John P. Stapp while stopping was approximately \( 20.63\, g \).

Key Concepts

Understanding Acceleration

Acceleration is all about how quickly an object changes its velocity. It's not just speeding up; it can be slowing down as well, which is called deceleration. When you hear acceleration, think of it like the "pedal to the metal" in a car, which makes you go faster. However, in physics, it’s about any change in speed or direction.
There are three ways to accelerate:

• Speeding up (positive acceleration)
• Slowing down (deceleration or negative acceleration)
• Changing direction (like turning a corner)

To find acceleration, we use the formula: \[ a = \frac{\Delta v}{\Delta t} \]Where \(\Delta v\) is the change in velocity, and \(\Delta t\) is the time over which this change happens. This formula helps us see exactly how fast your speed is changing over that time. Even stopping, like in the exercise with Colonel John P. Stapp on his rocket sled, involves a form of acceleration as the sled slows to a stop.

Conversion of Units Made Simple

When solving physics problems, it's often necessary to convert units to make calculations easier. In most physics problems, the standard unit for speed is meters per second (m/s), while the given speed may initially be in kilometers per hour (km/h). Here’s how we make that conversion.
The conversion factor from kilometers per hour to meters per second is that 1 km/h equals \(\frac{1}{3.6}\) m/s. This means:

• To convert km/h to m/s, multiply the speed by \(\frac{1}{3.6}\).
• In our exercise, \(1020 \text{ km/h} \times \frac{1}{3.6} = 283.33 \text{ m/s}\).

Knowing how to convert units is crucial for accuracy in physics as it ensures that the numbers you work with make sense in the context of the problem. It's like changing languages; different units tell what kind of "language" you need to speak for that particular part of the calculation.

Grasping Deceleration in Physics

Deceleration is simply acceleration in the opposite direction of motion, often referred to as negative acceleration. It’s what happens when an object slows down over time, and in the context of our exercise, it's what happened to Colonel Stapp's sled as it came to a stop.
When calculating deceleration, we still use the standard acceleration formula \(a = \frac{\Delta v}{\Delta t}\), but this time, \(\Delta v\) represents a decrease in velocity. The final result is thus a negative value indicating slowing down.
In terms of physics:

• Deceleration means the final velocity \(v_f\) is less than the initial velocity \(v_i\), and the velocity change \(\Delta v = v_f - v_i\) becomes negative.
• For Colonel Stapp's ride, the initial speed was 283.33 m/s, dropping to 0 m/s, so the change in velocity is negative.

Expressing the deceleration in terms of gravity (\(g\)), which is approximately \(9.81 \text{ m/s}^2\), makes it easier to comprehend how strong the force was - up to about \(20.63\) times stronger than regular gravitational pull! This helps visualize the immense force involved in bringing that speeding sled to a stop.