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Chapter 4: Problem 15

A particle leaves the origin with an initial velocity \(\vec{v}=(3.00 \hat{\mathrm{i}}) \mathrm{m} / \mathrm{s}\) and a constant acceleration \(\vec{a}=(-1.00 \hat{\mathrm{i}}-\) \(0.500 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2}\). When it reaches its maximum \(x\) coordinate, what are its (a) velocity and (b) position vector?

Short Answer

Expert verified
(a) \(\vec{v} = -1.50 \hat{\mathrm{j}} \, \text{m/s}\); (b) \(\vec{s} = 4.50 \hat{\mathrm{i}} - 2.25 \hat{\mathrm{j}} \, \text{m}\).

Step by step solution

01

Define maximum x coordinate condition

The maximum x-coordinate occurs when the x-component of the velocity is zero. Since the initial velocity in the x-direction is given by \(v_{x0} = 3.00 \, \text{m/s}\) and the acceleration in the x-direction is \(a_{x} = -1.00 \, \text{m/s}^2\), we set the final x-velocity \(v_{x}\) at the maximum position to zero.
02

Use kinematic equation for x-direction

The kinematic equation \(v_{x} = v_{x0} + a_{x} t\) can be used, where \(v_{x} = 0\) at the maximum x position. Substituting the values, we get:\[0 = 3.00 - 1.00 t\]Solving for \(t\), we find \(t = 3.00 \, \text{s}\).
03

Determine velocity at maximum x

The velocity at any point is given by \(\vec{v} = v_{x} \hat{\mathrm{i}} + v_{y} \hat{\mathrm{j}}\). At maximum x, \(v_{x} = 0\), and we need \(v_{y}\). Since \(v_{y0} = 0\) and \(a_{y} = -0.500 \, \text{m/s}^2\), using \(v_{y} = v_{y0} + a_{y} t\), substitute to find:\[v_{y} = 0 - 0.500 \times 3.00 = -1.50 \, \text{m/s}\]The velocity at maximum x coordinate is \(\vec{v} = -1.50 \hat{\mathrm{j}} \, \text{m/s}\).
04

Determine position vector at maximum x coordinate

The position vector \(\vec{s}\) is given by \(\vec{s} = s_{x} \hat{\mathrm{i}} + s_{y} \hat{\mathrm{j}}\). Use the equations for position: In x-direction: \[s_{x} = v_{x0} t + \frac{1}{2} a_{x} t^2 \]Substitute values:\[s_{x} = 3.00 \times 3.00 + \frac{1}{2} \times (-1.00) \times (3.00)^2 = 4.50 \, \text{m} \]In y-direction:\[s_{y} = v_{y0} t + \frac{1}{2} a_{y} t^2 \]Substitute values:\[s_{y} = 0 \times 3.00 + \frac{1}{2} (-0.500) (3.00)^2 = -2.25 \, \text{m} \]Thus, the position vector at maximum x coordinate is \(\vec{s} = 4.50 \hat{\mathrm{i}} - 2.25 \hat{\mathrm{j}} \, \text{m}\).

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Key Concepts

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

Particle Motion
Particle motion refers to the way in which a particle travels through space over time. In physics, we're often interested in understanding this movement since it's a fundamental part of kinematics. When we describe particle motion, we usually consider the particle's path and how its position changes.
  • The path a particle takes can be straight, curved, or a combination of both.
  • Describing the motion involves specifying the particle's position, velocity, and acceleration at any point in time.
Understanding particle motion is crucial because it allows us to predict the future position of a particle. By applying kinematic formulas, we can determine various attributes, such as velocity at different times, even if the particle undergoes constant acceleration.
Initial Velocity
Initial velocity is the velocity of a particle at the start of its motion. It's a key concept in understanding kinematics because it serves as the starting point for predicting future motion.
  • In our exercise, the initial velocity of the particle is given by \(\vec{v} = 3.00 \hat{\mathrm{i}}\, \, \mathrm{m/s}\).
  • This means that at \(t = 0\), the particle is moving at 3.00 m/s in the positive x-direction.
Initial velocity is crucial in the kinematic equations because it helps us determine how the motion develops over time. Knowing the initial velocity, along with acceleration, we can predict the particle's behavior, such as how far it will travel in a given time.
Constant Acceleration
Constant acceleration occurs when a particle's velocity changes at a steady rate over time. This is a fundamental idea in kinematics and simplifies the calculation of motion.
  • In the example provided, the particle has a constant acceleration \(\vec{a} = (-1.00 \hat{\mathrm{i}} - 0.500 \hat{\mathrm{j}})\, \mathrm{m/s}^2\).
  • This means the particle's speed decreases in the x-direction and increases in speed in the negative y-direction.
Constant acceleration allows us to use specific kinematic equations to calculate velocity and position as they relate to time. It's helpful because it provides a predictable framework to analyze the particle's motion efficiently, especially when identifying moments such as reaching a maximum x-coordinate.
Position Vector
The position vector indicates the position of a particle in a coordinate system, with a given magnitude and direction. It's crucial for tracking where the particle is at any given time.
  • The position vector in the exercise is given by \(\vec{s} = s_{x} \hat{\mathrm{i}} + s_{y} \hat{\mathrm{j}}\).
  • For our solution, at the maximum x-coordinate, the position vector is \(\vec{s} = 4.50 \hat{\mathrm{i}} - 2.25 \hat{\mathrm{j}} \mathrm{m}\).
The position vector not only tells us the final destination or location of the particle but also helps us visualize and comprehend its path travelled. It serves as a critical component in the kinematic analysis, helping us understand how far in each direction the particle has moved relative to its starting point.

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Most popular questions from this chapter

A baseball is hit at ground level. The ball reaches its maximum height above ground level \(3.0 \mathrm{~s}\) after being hit. Then \(2.5 \mathrm{~s}\) after reaching its maximum height, the ball barely clears a fence that is \(97.5 \mathrm{~m}\) from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) (uniform circular motion) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

A football kicker can give the ball an initial speed of \(25 \mathrm{~m} / \mathrm{s}\). What are the (a) least and (b) greatest elevation angles at which he can kick the ball to score a field goal from a point \(50 \mathrm{~m}\) in front of goalposts whose horizontal bar is \(3.44 \mathrm{~m}\) above the ground?

-1 The position vector for an electron is \(\vec{r}=(5.0 \mathrm{~m}) \hat{\mathrm{i}}-\) \((3.0 \mathrm{~m}) \hat{\mathrm{j}}+(2.0 \mathrm{~m}) \hat{\mathrm{k}}\). (a) Find the magnitude of \(\vec{r}\). (b) Sketch the vector on a right-handed coordinate system. -2 A watermelon seed has the following coordinates: \(x=-5.0 \mathrm{~m}\), \(y=8.0 \mathrm{~m}\), and \(z=0 \mathrm{~m} .\) Find its position vector (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the \(x\) axis. (d) Sketch the vector on a right-handed coordinate system. If the seed is moved to the \(x y z\) coordinates \((3.00 \mathrm{~m},\), \(0 \mathrm{~m}, 0 \mathrm{~m}\) ), what is its displacement (e) in unit-vector notation and as (f) a magnitude and (g) an angle relative to the positive \(x\) direction?

The fast French train known as the TGV (Train 脿 Grande Vitesse) has a scheduled average speed of \(216 \mathrm{~km} / \mathrm{h} .\) (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to \(0.050 \mathrm{~g}\), what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a \(1.00 \mathrm{~km}\) radius to be at the acceleration limit?
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