Solution
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Linear motion, also known as rectilinear motion, is a type of motion along a straight line. In the real world, there are several examples that can be modeled by linear motion. One of the most straightforward examples of linear motion is a car driving along a straight road at a constant speed.
Linear motion, also known as rectilinear motion, is a type of motion along a straight line. In the real world, there are several examples that can be modeled by linear motion. One of the most straightforward examples of linear motion is a car driving along a straight road at a constant speed.
Let's consider a car traveling on a straight highway. The car starts from rest and accelerates to a speed of 60 miles per hour (mph) in 10 seconds. It then maintains this speed for 2 minutes before decelerating to a stop in another 10 seconds. We will analyze this motion step by step.
Step 1: Acceleration Phase
The car starts from rest, which means its initial velocity ($v_i$) is 0 mph. It accelerates to 60 mph in 10 seconds. To convert the speed to feet per second (fps), we use the conversion factor 1 mph = 1.467 fps.
$$ v_f = 60 \text{ mph} \times 1.467 \text{ fps/mph} = 88.02 \text{ fps} $$
The acceleration ($a$) can be calculated using the formula:
$$ a = \frac{v_f - v_i}{t} $$
where $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the time taken to reach the final velocity.
$$ a = \frac{88.02 \text{ fps} - 0 \text{ fps}}{10 \text{ s}} = 8.802 \text{ fps}^2 $$
Step 2: Constant Speed Phase
During this phase, the car travels at a constant speed of 88.02 fps for 2 minutes (120 seconds). The distance ($d$) traveled can be calculated using the formula:
$$ d = v \times t $$
where $v$ is the velocity and $t$ is the time.
$$ d = 88.02 \text{ fps} \times 120 \text{ s} = 10,562.4 \text{ feet} $$
Step 3: Deceleration Phase
The car decelerates to a stop in 10 seconds. The deceleration ($a_d$) can be calculated using the same formula for acceleration, but since the car is coming to a stop, the final velocity will be 0 fps.
$$ a_d = \frac{0 \text{ fps} - 88.02 \text{ fps}}{10 \text{ s}} = -8.802 \text{ fps}^2 $$
The negative sign indicates that the car is slowing down. The distance covered during deceleration can be calculated using the kinematic equation:
$$ d = v_i t + \frac{1}{2} a_d t^2 $$
$$ d = 88.02 \text{ fps} \times 10 \text{ s} + \frac{1}{2} \times (-8.802 \text{ fps}^2) \times (10 \text{ s})^2 $$
$$ d = 880.2 \text{ feet} - 440.1 \text{ feet} = 440.1 \text{ feet} $$
Conclusion
In this real-world example, the car's motion can be divided into three distinct phases: acceleration, constant speed, and deceleration. Each phase can be described using the principles of linear motion, and the corresponding distances and velocities can be calculated using kinematic equations. This example demonstrates how linear motion can be applied to analyze and predict the behavior of objects moving along a straight path.
Let's consider a car traveling on a straight highway. The car starts from rest and accelerates to a speed of 60 miles per hour (mph) in 10 seconds. It then maintains this speed for 2 minutes before decelerating to a stop in another 10 seconds. We will analyze this motion step by step.
Step 1: Acceleration Phase
The car starts from rest, which means its initial velocity ($v_i$) is 0 mph. It accelerates to 60 mph in 10 seconds. To convert the speed to feet per second (fps), we use the conversion factor 1 mph = 1.467 fps.
$$ v_f = 60 \text{ mph} \times 1.467 \text{ fps/mph} = 88.02 \text{ fps} $$
The acceleration ($a$) can be calculated using the formula:
$$ a = \frac{v_f - v_i}{t} $$
where $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the time taken to reach the final velocity.
$$ a = \frac{88.02 \text{ fps} - 0 \text{ fps}}{10 \text{ s}} = 8.802 \text{ fps}^2 $$
Step 2: Constant Speed Phase
During this phase, the car travels at a constant speed of 88.02 fps for 2 minutes (120 seconds). The distance ($d$) traveled can be calculated using the formula:
$$ d = v \times t $$
where $v$ is the velocity and $t$ is the time.
$$ d = 88.02 \text{ fps} \times 120 \text{ s} = 10,562.4 \text{ feet} $$
Step 3: Deceleration Phase
The car decelerates to a stop in 10 seconds. The deceleration ($a_d$) can be calculated using the same formula for acceleration, but since the car is coming to a stop, the final velocity will be 0 fps.
$$ a_d = \frac{0 \text{ fps} - 88.02 \text{ fps}}{10 \text{ s}} = -8.802 \text{ fps}^2 $$
The negative sign indicates that the car is slowing down. The distance covered during deceleration can be calculated using the kinematic equation:
$$ d = v_i t + \frac{1}{2} a_d t^2 $$
$$ d = 88.02 \text{ fps} \times 10 \text{ s} + \frac{1}{2} \times (-8.802 \text{ fps}^2) \times (10 \text{ s})^2 $$
$$ d = 880.2 \text{ feet} - 440.1 \text{ feet} = 440.1 \text{ feet} $$
Conclusion
In this real-world example, the car's motion can be divided into three distinct phases: acceleration, constant speed, and deceleration. Each phase can be described using the principles of linear motion, and the corresponding distances and velocities can be calculated using kinematic equations. This example demonstrates how linear motion can be applied to analyze and predict the behavior of objects moving along a straight path.
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