Problem 77 - Fundamentals of Physics

Question

A hot rod can accelerate from 0 to \(60 \mathrm{~km} / \mathrm{h}\) in \(5.4 \mathrm{~s}\). (a) What is its average acceleration. in \(\mathrm{m} / \mathrm{s}^{2}\), during this time? (b) How far will it travel during the \(5.4 \mathrm{~s}\), assuming its acceleration is constant? (c) From rest, how much time would it require to go a distance of \(0.25 \mathrm{~km}\) if its acceleration could be maintained at the value in (a)?

Step-by-Step Solution

Answer

(a) 3.09 m/s², (b) 45 m, (c) 12.71 s.

Step 1: Convert Velocity to Meters per Second

First, we need to convert the final velocity from kilometers per hour to meters per second. We have a final velocity of 60 km/h. To convert it to m/s, use the conversion factor: \[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]Thus, \[ 60 \text{ km/h} \times \frac{1}{3.6} = \frac{60}{3.6} = 16.67 \text{ m/s} \]

Step 2: Calculate Average Acceleration

Now that we have the final velocity in meters per second, we can calculate the average acceleration. The initial velocity was 0 m/s, and the time taken was 5.4 seconds.The formula for average acceleration \( a \) is:\[ a = \frac{v_f - v_i}{t} \]Where:- \( v_f = 16.67 \text{ m/s} \) (final velocity)- \( v_i = 0 \text{ m/s} \) (initial velocity)- \( t = 5.4 \text{ s} \) (time)Substitute these values into the formula:\[ a = \frac{16.67 - 0}{5.4} \approx 3.09 \text{ m/s}^2 \]

Step 3: Calculate Distance Traveled with Constant Acceleration

To find the distance traveled during the 5.4 seconds with constant acceleration, we use the equation:\[ d = v_i t + \frac{1}{2} a t^2 \]Where:- \( v_i = 0 \text{ m/s} \)- \( a = 3.09 \text{ m/s}^2 \)- \( t = 5.4 \text{ s} \)Plug these values into the equation:\[ d = 0 \times 5.4 + \frac{1}{2} \times 3.09 \times (5.4)^2 \]\[ d = \frac{1}{2} \times 3.09 \times 29.16 \approx 45 \text{ meters} \]

Step 4: Calculate Time to Travel 0.25 km with Constant Acceleration

To find the time required to travel a distance of 0.25 km (or 250 m) with the same acceleration, we use the equation:\[ d = \frac{1}{2} a t^2 \]Rearrange for \( t \):\[ t = \sqrt{\frac{2d}{a}} \]Substitute \( d = 250 \text{ m} \) and \( a = 3.09 \text{ m/s}^2 \):\[ t = \sqrt{\frac{2 \times 250}{3.09}} \]\[ t \approx \sqrt{161.55} \approx 12.71 \text{ seconds} \]

Key Concepts

Constant Acceleration

In physics, constant acceleration implies that an object's velocity changes at a steady rate over time. This means that every second, the velocity increases or decreases by a fixed amount. If the acceleration is positive, the object's speed increases, while a negative acceleration (deceleration) leads to a decrease in speed. Constant acceleration makes the computation of various motion parameters straightforward and is frequently encountered in kinematic problems.

The hot rod problem here is an excellent example of constant acceleration. From rest, it accelerates uniformly up to 60 km/h in exactly 5.4 seconds. This allows us to apply consistent formulas to derive characteristics like average acceleration and distance traveled.

• Average acceleration, in cases of constant acceleration, helps us understand the change in velocity over specific intervals.
• It simplifies motion analysis by assuming steady changes, making it easier to predict future motion.

Kinematic Equations

Kinematic equations are powerful tools in physics that allow us to describe the motion of objects subject to constant acceleration. They link various quantities such as displacement, initial velocity, final velocity, acceleration, and time.

For our exercise, we used the kinematic equation \[ a = \frac{v_f - v_i}{t} \] to compute the average acceleration, where:

• \( v_f \): Final velocity
• \( v_i \): Initial velocity
• \( t \): Time interval

Additionally, the equation \[ d = v_i t + \frac{1}{2} a t^2 \] allowed us to calculate the distance traveled while the car was accelerating. This formula is essential to derive other motion parameters for various scenarios. It's crucial to understand how these equations can be rearranged to solve for different variables as needed, like solving for time when distance and acceleration are known.

• They enable analysis of how quickly an object speeds up or slows down.
• Through practice, understanding these equations becomes intuitive and immensely useful.

Velocity Conversion

Velocity conversion is a necessary step when solving kinematic problems, especially when given speed in units not directly compatible with the formulas. The most common conversion is from kilometers per hour (km/h) to meters per second (m/s).

In our exercise, the hot rod's final velocity needed conversion from 60 km/h to 16.67 m/s. This conversion is done using the factor:\[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]This ensures all units are consistent, so calculations like acceleration and distance are correct.

• Always check the units of velocity before performing calculations.
• Consistency with units is vital to avoid errors and ensure accuracy.

Distance Traveled

Calculating the distance traveled is a key part of understanding motion. When dealing with constant acceleration, the equation:\[ d = v_i t + \frac{1}{2} a t^2 \] is used. For our hot rod, the initial velocity \( v_i \) was zero, simplifying the equation to \( d = \frac{1}{2} a t^2 \).

In such scenarios, understanding how initial conditions affect distance is crucial. By knowing the acceleration and time, we were able to solve for the total distance during acceleration. For different initial conditions, the same equation applies, but it's important to assess each term's contribution based on initial values.

• The formula allows prediction of future positioning when the acceleration is constant.
• Analyzing distance traveled helps illustrate how quickly an object covers ground with varying acceleration rates.