Problem 107 - Fundamentals of Physics

Question

The head of a rattlesnake can accelerate at \(50 \mathrm{~m} / \mathrm{s}^{2}\) in striking a victim. If a car could do as well, how long would it take to reach a speed of \(100 \mathrm{~km} / \mathrm{h}\) from rest?

Step-by-Step Solution

Answer

It would take approximately 0.556 seconds to reach 100 km/h.

Step 1: Convert Units

The final speed is given in kilometers per hour (km/h). We need to convert this to meters per second (m/s) for consistency with the given acceleration. Use the conversion factor: \[1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}.\]Thus, \[100 \text{ km/h} = \frac{100}{3.6} \text{ m/s} \approx 27.78 \text{ m/s}.\]

Step 2: Use the Equation of Motion

Using the formula for acceleration, which is given by \[ v = u + at,\]where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. Since the car starts from rest, \(u = 0\). The formula simplifies to:\[ v = at.\]

Step 3: Solve for Time

Substitute \(v = 27.78 \text{ m/s}\) and \(a = 50 \text{ m/s}^2\) into the equation \[ v = at \].Solve for \(t\): \[ t = \frac{v}{a} = \frac{27.78}{50} \approx 0.556 \text{ seconds}.\]

Key Concepts

acceleration

Acceleration is the rate at which an object changes its velocity. It tells us how quickly something is speeding up or slowing down. In physics, acceleration is a vector quantity, which means it has both a magnitude and a direction.

When we talk about acceleration in kinematics, we're often referring to linear acceleration, meaning the change in velocity in a straight line. The unit of acceleration in the International System of Units (SI) is meters per second squared (\(\mathrm{m/s^{2}} \)). This unit indicates how much the velocity of an object increases every second.

For example, in the exercise, the rattlesnake's head can accelerate at 50 \(\mathrm{m/s^{2}} \). This means if it started from rest, its velocity would increase by 50 meters per second (m/s) every second. Understanding this concept is crucial when calculating how long it takes for an object to reach a certain speed.

velocity

Velocity describes the speed of an object in a particular direction. Unlike speed, which is scalar and only provides magnitude, velocity is a vector. This means it provides both the rate of motion and the direction in which the object is moving.

In physics problems, like the one provided, velocity is often given in meters per second (\(\mathrm{m/s} \)). However, other units such as kilometers per hour (km/h) can also be used. To accurately compute problems involving kinematics, you may need to convert between these units.

The initial velocity (denoted as \(u\)) is the velocity at the start of the observation, and the final velocity (\(v\)) is the velocity at the end. In the exercise, we start from rest, so our initial velocity \(u\) is 0. We need to find how long it takes to reach a final velocity of approximately 27.78 m/s, which is the converted speed from 100 km/h.

equation of motion

The equations of motion are essential tools in kinematics. They help us understand how objects move under various forces and conditions. These equations relate displacement, velocity, acceleration, and time.

One of the most commonly used equations of motion is:\[v = u + at\]Here:

• \(v\) is the final velocity,
• \(u\) is the initial velocity,
• \(a\) is the acceleration,
• \(t\) is the time.

This equation becomes particularly useful when initial velocity \(u\) is zero, simplifying to \(v = at\). This allows us to calculate the time \(t\) it takes to reach a desired velocity if the acceleration is known.

In the exercise noted, we use this equation to determine how long it takes for the car to reach the velocity of 27.78 m/s with an acceleration of 50 m/s².

unit conversion

Unit conversion is the process of converting quantities from one unit to another. It's fundamental in physics to ensure consistency and accuracy in solving problems.

In kinematics, especially, it's common to convert speeds from kilometers per hour (km/h) to meters per second (m/s) because acceleration is usually given in m/s².

For example, to convert 100 km/h to m/s, you use the conversion factor:\[1 \, \mathrm{km/h} = \frac{1}{3.6} \, \mathrm{m/s}\]Thus, converting 100 km/h:\[100 \, \mathrm{km/h} = \frac{100}{3.6} \, \mathrm{m/s} \approx 27.78 \, \mathrm{m/s}\]
Without correctly converting units, inaccuracies may arise in calculations. Proper unit conversion is critical for ensuring that all aspects of a physics problem align and lead to correct results.