Question

A jet touches down on a runway with a speed of \(142.4 \mathrm{mph} .\) After \(12.4 \mathrm{~s},\) the jet comes to a complete stop. Assuming constant acceleration of the jet, how far down the runway from where it touched down does the jet stand?

Short answer

Answer: The jet stands about 787.44 meters down the runway after decelerating to a complete stop.

Step-by-step solution

Identify the known values

The problem provides the following information: - Initial velocity of the jet, \(v_i = 142.4\,\mathrm{mph}\) - Time of acceleration, \(t = 12.4\,\mathrm{s}\) - Final velocity, \(v_f = 0\) since the jet comes to a complete stop.

Convert the initial velocity to \(\mathrm{m/s}\)

Since we have the time in seconds, we need to convert the initial velocity from \(\mathrm{mph}\) to \(\mathrm{m/s}\): $$v_i = 142.4\,\mathrm{mph} \cdot 1609.34 \mathrm{m/mile} \cdot \dfrac{1\,\mathrm{h}}{3600\,\mathrm{s}} ≈ 63.6\,\mathrm{m/s}$$

Find the acceleration

Using the formula \(v_f = v_i + at\), we can solve for acceleration \(a\): $$a = \dfrac{v_f - v_i}{t} = \dfrac{0 - 63.6\,\mathrm{m/s}}{12.4\,\mathrm{s}} ≈ -5.14\,\mathrm{m/s^2}$$

Find the distance traveled

Finally, use the equation \(d = v_i t + 0.5at^2\) to find the distance the jet traveled on the runway: $$d = 63.6\,\mathrm{m/s}\cdot 12.4\,\mathrm{s} + 0.5\cdot (-5.14\,\mathrm{m/s^2})(12.4\,\mathrm{s})^2 ≈ 787.44\,\mathrm{m}$$ So, the jet stands about \(787.44\,\mathrm{m}\) down from its touchdown on the runway.

Key concepts

Constant Acceleration

In physics, constant acceleration refers to a situation where the speed of an object changes at a steady rate over time. When a jet lands on a runway, if it decelerates at a constant rate, its acceleration is negative because it is slowing down. Constant acceleration is a key concept because it simplifies calculations, enabling us to predict future positions and velocities easily.

With constant acceleration, the formula \(v_f = v_i + at\) is frequently used, where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. This formula helps in determining the rate at which the speed reduces as the jet halts. In our example, the jet's deceleration from 63.6 m/s to 0 m/s over 12.4 seconds illustrates this principle.

Using this steady rate of speed change allows us to further calculate how far the jet traveled down the runway during deceleration using the distance formula for constant acceleration: \(d = v_i t + 0.5 a t^2\). This predictable pattern is incredibly useful in various real-world physics applications.

Kinematics

Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause the motion. It heavily relies on quantities such as velocity, acceleration, time, and distance. In our jet landing scenario, kinematics principles help us capture the motion characteristics through equations that correlate these quantities.

The key kinematics equations used are:

• \(v_f = v_i + at\): Relates final and initial velocity, acceleration, and time.
• \(d = v_i t + 0.5 a t^2\): Determines the distance covered under constant acceleration.

In this example, we start with the jet's known initial speed and the time it takes to stop. We use the first equation to find the acceleration and the second to calculate the distance traveled until it comes to rest.

Kinematics allows us to predict the future state of any moving object provided we have initial conditions and assume uniform motion characteristics, simplifying the often complex nature of analyzing motion.

Unit Conversion

Unit conversion is critical in physics to ensure calculations are accurately performed by using consistent units. In our jet's example, the speed begins in miles per hour (mph), which needs conversion to meters per second (m/s) because time is given in seconds, and standard physics formulas use meters and seconds for simplicity and universality.

The conversion from mph to m/s requires two steps:

• Convert miles to meters: Multiply the miles by 1,609.34, since there are 1,609.34 meters in a mile.
• Convert hours to seconds: Multiply the resulting value by \(\frac{1}{3600}\), as there are 3,600 seconds in an hour.

This process gives us the initial speed in m/s, which is approximately 63.6 m/s.

Consistent unit usage is necessary to avoid errors in calculations—especially in physics problems—where mixing units can result in confusion or incorrect answers. Understanding and performing unit conversions is a skill set that ensures precise problem-solving across various contexts.