Problem 27 - College Physics

Question

Cheetahs, the fastest of the great cats, can reach \(45 \mathrm{mi} / \mathrm{h}\) in \(2.0 \mathrm{~s}\) starting from rest. Assuming that they have constant acceleration throughout that time, find (a) their acceleration (in \(\mathrm{ft} / \mathrm{s}^{2}\) and \(\left.\mathrm{m} / \mathrm{s}^{2}\right)\) and \((\mathrm{b})\) the distance (in \(\mathrm{m}\) and \(\mathrm{ft}\) ) they travel during that time.

Step-by-Step Solution

Answer

Acceleration: 10.06 m/s² or 33 ft/s². Distance: 20.1 m or 66 ft.

Step 1: Convert Units of Velocity

First, convert the final velocity from miles per hour to meters per second and feet per second. 1 mile = 1609.344 meters and 1 hour = 3600 seconds.Thus, 45 mi/h to m/s is given by:\[45 \frac{mi}{h} \times \frac{1609.344 \text{ meters}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} \approx 20.1168 \frac{m}{s}\]Similarly, to find 45 mi/h in ft/s:1 mile = 5280 feet, hence:\[45 \frac{mi}{h} \times \frac{5280 \text{ feet}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} \approx 66 \frac{ft}{s}\]

Step 2: Calculate Acceleration

Since the cheetah starts from rest, the initial velocity \(u = 0\). The formula for acceleration \(a\) when we know the final velocity \(v\), initial velocity \(u\), and time \(t\) is:\[ a = \frac{v-u}{t}\]Substitute the values in meters per second:\[ a = \frac{20.1168 \frac{m}{s} - 0}{2 \text{ s}} = 10.0584 \frac{m}{s^2}\]Now substitute the values in feet per second:\[ a = \frac{66 \frac{ft}{s} - 0}{2 \text{ s}} = 33 \frac{ft}{s^2}\]

Step 3: Calculate Distance Traveled in Feet and Meters

Use the formula for distance \(s\) traveled under constant acceleration which is given by:\[ s = ut + \frac{1}{2}at^2\]With \(u = 0\), this simplifies to:\[ s = \frac{1}{2}a t^2\]First, calculate in meters:\[ s = \frac{1}{2} \times 10.0584 \frac{m}{s^2} \times (2 \text{ s})^2 = 20.1168 \text{ meters}\]Now, calculate in feet:\[ s = \frac{1}{2} \times 33 \frac{ft}{s^2} \times (2 \text{ s})^2 = 66 \text{ feet}\]

Key Concepts

Unit Conversion in Physics

In physics, unit conversion is crucial for proper analysis and understanding. It involves converting various measurements so they can be used consistently within equations and calculations. For example, in the exercise, we needed to convert a velocity of 45 miles per hour (mi/h) into meters per second (m/s) and feet per second (ft/s). This is because different formulas may require different units depending on the context.

• To convert miles per hour to meters per second, you multiply by the number of meters in a mile and divide by the number of seconds in an hour.
• To convert miles per hour to feet per second, multiply by the number of feet in a mile and divide by the seconds in an hour.

Converting units allows us to apply physics equations correctly and compare results accurately, hence it is an essential skill in physics.

Constant Acceleration Motion

Constant acceleration motion is a scenario in which an object's acceleration does not change over time. It’s a core concept in kinematics and helps in predicting the object's future motion.

• Acceleration is the rate of change of velocity per unit of time.
• For constant acceleration, the rate stays stable, making calculations more straightforward.

In our cheetah example, when starting from rest and accelerating steadily to its top speed, the acceleration remains the same throughout. This simplification allows us to use straightforward formulas to calculate properties like acceleration and distance. Understanding constant acceleration is beneficial as it simplifies complex real-world motion into predictable outcomes.

Distance and Velocity Calculations

Calculating distance and velocity correctly requires knowing the initial conditions such as starting velocity (u) and acceleration (a), as well as the time (t) over which acceleration occurs. The formula used here for distance is derived from basic motion equations.

• The velocity of an object tells you how fast it's moving in a particular direction.
• By incorporating time, we can calculate how far it has traveled under constant acceleration using \[ s = ut + \frac{1}{2} a t^2 \] where \(s\) is the distance, \(u\) is the initial velocity, \(a\) is acceleration, and \(t\) is time.

For the cheetah's case, we used the simplified form \( s = \frac{1}{2} a t^2 \) since the initial velocity is zero. Calculating distance helps us understand the extent of movement during accelerated motion.

Cheetah Speed Analysis

Analyzing cheetah speed offers valuable insight into nature's incredible designs and draws attention to real-world physics applications. Cheating physics, cheetahs can accelerate up to 45 mi/h in a mere 2.0 seconds.

• This rapid acceleration is a demonstration of strong muscular force acting over a short time.
• Understanding this acceleration helps us realize how physics principles apply not just in the laboratory but in everyday life.
  

By applying constant acceleration formulas, we learned how rapidly the cheetah accelerates, the ensuing speeds, and how far it travels during high-speed bursts. It's a lesson in how natural abilities can be quantitatively analyzed to better appreciate and understand biological machines.

Physics in Real-world Applications

Physics isn't just theoretical; it has numerous real-world applications, aiding us in understanding and solving actual problems—like how a cheetah sprints.

• Everyday examples of similar principles include the mathematics of car acceleration, where velocity and distance matter.
• Engineers develop better transportation systems by applying the understanding of acceleration and velocity.

Through studying motion, distance, and speed, we improve technologies and enhance knowledge about both human and animal efficiency.
Using physics principles practically helps us make innovations that improve life, from better vehicles to understanding wildlife. Thus, physics offers profound insights into both our immediate and broader environments.