91Ó°ÊÓ

A truck's speed increases uniformly from \(15 \mathrm{~km} / \mathrm{h}\) to \(60 \mathrm{~km} / \mathrm{h}\) in 20 s. Determine \((a)\) the average speed, \((b)\) the acceleration, and \((c)\) the distance traveled, all in units of meters and seconds. For the 20-s trip under discussion, taking \(+x\) to be in the direction of motion, $$ \begin{array}{l} v_{i x}=\left(15 \frac{\mathrm{k} m}{\not \mathrm{h}}\right)\left(1000 \frac{\mathrm{m}}{\mathrm{k} \not \mathrm{\gamma}}\right)\left(\frac{1}{3600} \frac{\not \mathrm{K}}{\mathrm{s}}\right)=4.17 \mathrm{~m} / \mathrm{s} \\ v_{f x}=60 \mathrm{~km} / \mathrm{h}=16.7 \mathrm{~m} / \mathrm{s} \end{array} $$ (a) \(v_{a v}=\frac{1}{2}\left(v_{i x}+v_{f x}\right)=\frac{1}{2}(4.17+16.7) \mathrm{m} / \mathrm{s}=10 \mathrm{~m} / \mathrm{s}\) (b) \(\quad a=\frac{v_{f x}-v_{i x}}{t}=\frac{(16.7-4.17) \mathrm{m} / \mathrm{s}}{20 \mathrm{~s}}=0.63 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(\quad x=v_{\text {av }} t=(10.4 \mathrm{~m} / \mathrm{s})(20 \mathrm{~s})=208 \mathrm{~m}=0.21 \mathrm{~km}\)

Step by step solution

01

Convert Speeds to Meters per Second

Start by converting the initial and final speeds from kilometers per hour (km/h) to meters per second (m/s). We use the conversion factor: 1 km/h = \( \frac{1000}{3600} \) m/s. Given: \(v_{ix} = 15 \frac{\mathrm{km}}{\mathrm{h}} \), convert to m/s: \( v_{ix} = 15 \times \frac{1000}{3600} = 4.17 \frac{\mathrm{m}}{\mathrm{s}} \).Similarly, for \(v_{fx} = 60 \frac{\mathrm{km}}{\mathrm{h}} \): \( v_{fx} = 60 \times \frac{1000}{3600} = 16.7 \frac{\mathrm{m}}{\mathrm{s}} \).

02

Calculate the Average Speed

The average speed \(v_{av}\) during uniform acceleration can be calculated as the mean of the initial and final speeds:\[v_{av} = \frac{1}{2}(v_{ix} + v_{fx})\]Substituting the values:\[v_{av} = \frac{1}{2}(4.17 + 16.7) = 10.4 \: \mathrm{m/s}\]

03

Calculate the Acceleration

Acceleration \(a\) is calculated using the formula for acceleration in uniform motion:\[a = \frac{v_{fx} - v_{ix}}{t}\]Where: - \(v_{fx} = 16.7 \) m/s- \(v_{ix} = 4.17 \) m/s- \(t = 20 \) sSubstituting the values:\[a = \frac{16.7 - 4.17}{20} = 0.63 \: \mathrm{m/s}^2\]

04

Calculate the Distance Traveled

The distance \(x\) traveled can be found using the formula:\[x = v_{av} \times t\]Substitute the values for \(v_{av}\) and time:\[x = 10.4 \times 20 = 208 \: \mathrm{m}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics

Kinematics is a branch of physics that deals with the motion of objects without considering the forces causing such motion. It focuses on the movement of objects using velocity, acceleration, displacement, and time. Kinematics is part of the foundation for understanding dynamics and is key in analyzing the motion depicted in our exercise.

When dealing with uniform acceleration, like in the given problem, the velocity of an object changes at a consistent rate. This means that the acceleration value remains constant over time. Understanding this concept is crucial for solving problems related to the motion of objects as it allows us to apply uniform kinematic equations.

• Initial velocity ( v_i )
• Final velocity ( v_f )
• Time ( t )
• Acceleration ( a )

These parameters are typically involved in the calculations to predict how far an object has moved or how long it will take to reach a certain velocity.

Conversion of Units

Conversion of units is an essential mathematical skill required to solve physics problems accurately. In this exercise, we are converting speeds from kilometers per hour ( ext{km/h} ) to meters per second ( ext{m/s} ). This is a common conversion because many physics equations use the SI unit of measure, which is meters per second for speed.

The conversion factor between ext{km/h} and ext{m/s} is calculated as follows:

• 1 kilometer equals 1000 meters.
• 1 hour equals 3600 seconds.

So the conversion factor is: \[ 1 ext{ km/h} = rac{1000}{3600} ext{ m/s} \].

By applying this conversion factor, we can convert the initial speed of 15 ext{km/h} to 4.17 ext{m/s} and the final speed of 60 ext{km/h} to 16.7 ext{m/s} . Such conversions allow us to use consistent units throughout the calculations.

Average Speed

Average speed refers to the total distance traveled divided by the total time taken. In kinematic problems involving uniform acceleration, it can also be calculated as the mean of the initial and final speeds. This is useful because it simplifies the expression when calculating the distance traveled.

In this problem, the formula used is:
\[ v_{av} = \frac{1}{2}(v_{ix} + v_{fx})\]
where:

• \( v_{ix} \) is the initial speed (4.17 ext{m/s} ) and
• \( v_{fx} \) is the final speed (16.7 ext{m/s} ).

This results in an average speed of 10 ext{m/s} . Average speed simplifies the process when calculating how much distance has been covered over the time interval.

Distance Calculation

Distance calculation in kinematics often involves using the average speed when acceleration is uniform. With the given uniform acceleration, the distance traveled can be determined by multiplying the average speed by the time duration.
The formula for distance ( x ) is:
\[ x = v_{av} imes t\]
In this exercise, we substitute the average speed of 10.4 ext{m/s} and the time of 20 seconds:
\[ x = 10.4 imes 20 = 208 ext{ meters}\]
Understanding how to calculate distance when the speed changes uniformly over time helps in solving a variety of real-world physics problems, such as predicting travel distances and times in transportation scenarios. The use of consistent units and the understanding of uniform acceleration streamlines the problem-solving process.