Question

A particle starts from rest at \(x=0\) and moves for \(20 \mathrm{~s}\) with an acceleration of \(+2.0 \mathrm{~cm} / \mathrm{s}^{2}\). For the next \(40 \mathrm{~s}\), the acceleration of the particle is \(-4.0 \mathrm{~cm} / \mathrm{s}^{2} .\) What is the position of the particle at the end of this motion?

Short answer

Answer: The final position (displacement) of the particle at the end of its motion is -1200 cm.

Step-by-step solution

Find the final velocity at the end of the first stage

In this step, we will use the formula \(v = v_0 + at\) to find the final velocity at the end of the first stage, where \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. The initial velocity \(v_0\) is 0 because the particle starts from rest, the acceleration \(a\) is 2.0 cm/s², and the time \(t\) is 20s. \(v = 0 + (2.0 \mathrm{~cm/s^2})(20 \mathrm{~s}) = 40\mathrm{~cm/s}\) The final velocity at the end of the first stage is 40 cm/s.

Calculate the displacement at the end of the first stage

Now we will use the formula \(x = x_0 + v_0t + \frac{1}{2}at^2\) to find the displacement at the end of the first stage. The initial position \(x_0\) is 0, the initial velocity \(v_0\) is 0, the acceleration \(a\) is 2.0 cm/s², and the time \(t\) is 20s. \(x = 0 + 0 + \frac{1}{2}(2.0 \mathrm{~cm/s^2})(20 \mathrm{~s})^2 = 400\mathrm{~cm}\) The displacement at the end of the first stage is 400 cm.

Find the final velocity at the end of the second stage

At the start of the second 40s interval, the velocity of the particle is \(40\mathrm{~cm/s}\) which we calculated earlier. We can use the formula \(v = v_0 + at\) to find the final velocity at the end of the second stage, where \(v_0\) is the initial velocity (40 cm/s), \(a\) is the acceleration (-4.0 cm/s²), and \(t\) is the time (40s). \(v = 40\mathrm{~cm/s} + (-4.0\mathrm{~cm/s^2})(40\mathrm{~s}) = -120\mathrm{~cm/s}\) The final velocity at the end of the second stage is -120 cm/s.

Calculate the displacement at the end of the second stage

Now we will use the formula \(x = x_0 + v_0t + \frac{1}{2}at^2\) to find the displacement at the end of the second stage. The initial position \(x_0\) is 0, the initial velocity \(v_0\) is 40 cm/s, the acceleration \(a\) is -4.0 cm/s², and the time \(t\) is 40s. \(x = 0 + (40\mathrm{~cm/s})(40\mathrm{~s}) + \frac{1}{2}(-4.0 \mathrm{~cm/s^2})(40 \mathrm{~s})^2 = 1600 - 3200 = -1600\mathrm{~cm}\) The displacement at the end of the second stage is -1600 cm.

Calculate the total displacement

Finally, we will add the displacement of the first stage (400 cm) and the displacement of the second stage (-1600 cm) to find the total displacement. Total displacement = Displacement of the first stage + Displacement of the second stage Total displacement = 400 cm + (-1600 cm) = -1200 cm The final position of the particle at the end of this motion is -1200 cm.

Key concepts

Uniform motion

Uniform motion refers to motion at a constant speed in a straight line. In situations of uniform motion, an object moves equal distances in equal time intervals.
In this problem, the particle does not experience uniform motion throughout its journey since its acceleration varies. However, understanding uniform motion lays the foundation for grasping more complex motion, such as the particle's path.
Key characteristics of uniform motion include:

• Constant velocity: No change in speed or direction.
• No acceleration: The velocity remains steady over time.

The importance of uniform motion in kinematics is fundamental since it forms the simplest case, leading us to analyze more complex scenarios where objects accelerate or decelerate. In this specific exercise, the particle first accelerates to reach a certain velocity, and later, experiences a different acceleration.

Displacement calculation

Displacement is the change in the position of an object, and it is a vector quantity, meaning it has both magnitude and direction.
To calculate displacement, the formula used is:\[x = x_0 + v_0t + \frac{1}{2}at^2\]Where:

• \(x_0\) is the initial position.
• \(v_0\) is the initial velocity.
• \(a\) is the acceleration.
• \(t\) is the time.

In the scenario presented, the particle undergoes two stages of movement:
The first stage: The particle starts from rest and has a positive acceleration, resulting in a displacement of 400 cm.
The second stage: The particle continues with an initial velocity from the first stage, but it experiences negative acceleration, which causes it to slow down, resulting in a displacement of -1600 cm.
In order to find the total displacement, we sum the displacements from each stage: 400 cm + (-1600 cm) = -1200 cm. The negative sign indicates the overall direction of the position change.

Acceleration

Acceleration is the rate at which an object's velocity changes with time. It is calculated using the formula:\[a = \frac{v - v_0}{t}\]Where:

• \(a\) is acceleration.
• \(v\) is the final velocity.
• \(v_0\) is the initial velocity.
• \(t\) is the time.

In this exercise, the acceleration is provided for two intervals. In the first interval, the particle starts at rest with an acceleration of 2.0 cm/s², leading to a final velocity of 40 cm/s. During the second interval, the particle experiences an acceleration of -4.0 cm/s², causing its velocity to decrease until it reaches -120 cm/s.
The key points to remember about acceleration:

• It is a vector quantity, which means it has both a magnitude and a direction.
• Positive acceleration results in an increase in velocity, while negative acceleration (also known as deceleration) results in a decrease in velocity, as shown in the exercise.

Understanding acceleration is crucial for analyzing motion since it helps describe how quickly an object can speed up or slow down.