Question

In the problems, please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. While passing a slower car on the highway, you accelerate uniformly from \(17.4 \mathrm{m} / \mathrm{s}\) to \(27.3 \mathrm{m} / \mathrm{s}\) in a time of \(10.0 \mathrm{s} .\) (a) How far do you travel during this time? (b) What is your acceleration magnitude?

Short answer

(a) The distance traveled is 223.5 m. (b) The acceleration is 0.99 m/s².

Step-by-step solution

Identify Given Values

Initial velocity, \( v_i = 17.4 \, \mathrm{m/s} \); final velocity, \( v_f = 27.3 \, \mathrm{m/s} \); time, \( t = 10.0 \, \mathrm{s} \). We need to find the distance traveled and the acceleration.

Calculate Acceleration

Use the formula for acceleration: \( a = \frac{v_f - v_i}{t} \). Substitute the given values: \( a = \frac{27.3 \, \mathrm{m/s} - 17.4 \, \mathrm{m/s}}{10.0 \, \mathrm{s}} = 0.99 \, \mathrm{m/s^2} \).

Calculate Distance Traveled

Use the kinematic equation for distance: \( d = v_i t + \frac{1}{2} a t^2 \). Substitute the known values: \( d = 17.4 \, \mathrm{m/s} \times 10.0 \, \mathrm{s} + \frac{1}{2} \times 0.99 \, \mathrm{m/s^2} \times (10.0 \,\mathrm{s})^2 \). Calculate \( d = 174 \, \mathrm{m} + 49.5 \, \mathrm{m} = 223.5 \, \mathrm{m} \).

Key concepts

Uniform Acceleration

Uniform acceleration occurs when an object's velocity changes at a constant rate over time. This means that every second, the object's speed increases by a consistent amount. This concept is crucial in understanding many real-world motions, such as a car smoothly speeding up from a stoplight. When acceleration is uniform, you can confidently predict the object's future conditions given that you know its current state.
In our exercise, the car exhibits uniform acceleration as it increases its speed from 17.4 m/s to 27.3 m/s over 10 seconds. This consistent change makes it easier to calculate both its acceleration and the distance it travels during this time. Recognizing uniform acceleration is essential when simplifying problems involving motion, because it allows us to use kinematic equations, which are tailor-made for these conditions.
When dealing with uniform acceleration, remember:

• Acceleration remains the same throughout the motion.
• The velocity changes linearly, which simplifies our calculations.

Kinematic Equations

Kinematic equations are mathematical expressions that describe the motion of objects undergoing uniform acceleration. These equations relate parameters such as velocity, time, displacement, and acceleration. By using kinematic equations, we can solve for unknown variables when an object's motion is characterized by constant acceleration. They provide a great toolkit for analyzing motion.
In solving our problem, we used the following kinematic equations:

• To find acceleration: \[ a = \frac{v_f - v_i}{t} \] where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is the time elapsed.
• To calculate distance: \[ d = v_i t + \frac{1}{2} a t^2 \] where \( d \) is distance, \( v_i \) is initial velocity, \( a \) is acceleration, and \( t \) is time.

These formulas simplify the relationships between physical quantities, enabling us to predict how objects behave when they move.

Distance Calculation

Calculating the distance traveled by an object is an important skill in understanding motion. In scenarios with uniform acceleration, we can efficiently determine the distance by employing the right kinematic equation.
In our exercise, we identified the initial velocity (\( 17.4 \mathrm{m/s} \)) and the time (10 seconds) to help calculate the distance traveled by the car while it accelerated uniformly. By applying the kinematic equation \( d = v_i t + \frac{1}{2} a t^2 \), we were able to break down the process:

• Firstly, compute distance contributed by initial velocity: \( v_i \times t = 17.4 \, \mathrm{m/s} \times 10 \, \mathrm{s} = 174 \, \mathrm{m} \).
• Next, calculate distance due to acceleration: \( \frac{1}{2} \times 0.99 \, \mathrm{m/s^2} \times 100 \, \mathrm{s^2} = 49.5 \, \mathrm{m} \).
• Finally, add these two results to find the total distance: \( 174 \, \mathrm{m} + 49.5 \, \mathrm{m} = 223.5 \, \mathrm{m} \).

Using the calculated total, we can understand how far the car moved during the acceleration period. Ensuring each step is understood can enhance problem-solving skills.

Acceleration Calculation

Finding acceleration is often a foundational step in solving motion problems, especially when uniform acceleration is involved. Acceleration quantifies how quickly velocity changes. It can be found using the basic formula \( a = \frac{v_f - v_i}{t} \).
In our problem, we calculated the car's acceleration while it overtook another on the highway. The initial velocity was 17.4 m/s and the final velocity was 27.3 m/s, over a time span of 10 seconds. So, the calculation was:

• Subtract the initial velocity from the final velocity: \( 27.3 \, \mathrm{m/s} - 17.4 \, \mathrm{m/s} = 9.9 \, \mathrm{m/s} \).
• Divide this change in velocity by the time period: \( \frac{9.9 \, \mathrm{m/s}}{10 \, \mathrm{s}} = 0.99 \, \mathrm{m/s^2} \).

The calculated acceleration (0.99 \( \mathrm{m/s^2} \)) is key to understanding the nature of motion experienced. Recognizing how acceleration impacts motion allows for comprehensive analysis of dynamics in motion.