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Chapter 4: Problem 21

A particle starts from the origin at \(t=0 \mathrm{~s}\) with a velocity of \(10.0 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}\) and moves \(\mathrm{in}\) the \(x\) - \(y\) plane with a constant acceleration of \((8.0 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}}) \mathrm{m} \mathrm{s}^{-2} .\) (a) At what time is the \(x\) - coordinate of the particle \(16 \mathrm{~m} ?\) What is the \(y\) -coordinate of the particle at that time? (b) What is the speed of the particle at the time?

Short Answer

Expert verified
At \(t = 2\) s, the y-coordinate is 22 m and the speed is approximately 21.3 m/s.

Step by step solution

01

Use the Kinematic Equation for x-coordinate

We use the kinematic equation for position in the x-direction: \( x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \). Since the particle starts at the origin, \(x_0 = 0\) and \(v_{0x} = 0\), so \( x(t) = \frac{1}{2} \times 8.0 \times t^2 \). Set this equal to 16 m, so: \( 16 = 4t^2 \).
02

Solve for Time when x = 16 m

Solving the equation from Step 1: \( 16 = 4t^2 \) leads to \( t^2 = 4 \). Taking the square root gives \( t = 2 \) s.
03

Find the y-coordinate at t = 2 s

Use the kinematic equation for position in the y-direction: \( y(t) = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \). With \(y_0 = 0\), \(v_{0y} = 10\) m/s, and \(a_y = 2\) m/sderiving \( y(t) = 10t + \frac{1}{2} \times 2 \times t^2\). For \(t = 2\) s, calculate \(y(2) = 10\times2 + 2 = 22\) m.
04

Determine the Speed at t = 2 s

First, find the velocity components: \(v_x = v_{0x} + a_x t = 0 + 8\times2 = 16\) m/s, \(v_y = v_{0y} + a_y t = 10 + 2\times2 = 14\) m/s. The speed is the magnitude of the velocity vector: \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{16^2 + 14^2} = \sqrt{256 + 196} = \sqrt{452} \approx 21.3\) m/s.

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Key Concepts

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

Constant Acceleration
In kinematics, constant acceleration means that an object's velocity changes by the same amount every second. When acceleration is constant, the kinematic equations can help predict the future positions and velocities of moving objects. These equations relate time, velocity, position, and acceleration.

For a particle moving with a constant acceleration like
  • Initial velocity: the velocity at which an object starts its motion, given here as \(v_0 = 10.0 \hat{\mathbf{j}} \text{ m/s} \).
  • Constant acceleration: influences how quickly an object speeds up or slows down over time. In this exercise, we see \(\mathbf{a} = 8.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}} \text{ m/s}^2\). Each component affects the particle's velocity along the respective axis.
  • Time (\(t\)): the variable we solve for when determining how long it takes for an object to reach a particular position.
With these values, we use kinematic equations to predict when and where the object will be at given times. The acceleration components influence the velocity in each direction independently.

Understanding constant acceleration is crucial in predicting the particle's path in motion scenarios like this one.
Velocity Calculation
Velocity is a vector quantity that specifies both the speed and direction of a moving object. Here, it's denoted with vectors because the particle travels in a 2D plane.

To calculate velocity with constant acceleration, these factors are essential:
  • Initial velocity (\(v_0\)): the starting speed, which in this problem for the y-direction is \(10 \text{ m/s} \), and \(0 \text{ m/s} \) along the x-direction.
  • Constant acceleration (\(\mathbf{a} \)): affects how the velocity changes over time, with effects seen along the x-axis and y-axis.
  • Time (\(t\)): determines how long the acceleration acts on the initial velocity.
To find the velocity at any moment, use the equation:\[v = v_0 + a\cdot t\]

In the particle's case, you have:
  • \(v_x = 0 + 8 \cdot 2 = 16 \text{ m/s}\)
  • \(v_y = 10 + 2 \cdot 2 = 14 \text{ m/s}\)
These help in finding the overall speed of the particle, using Pythagoras to calculate the magnitude:\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{16^2 + 14^2} \approx 21.3 \text{ m/s}\]

This final speed combines the velocities in both directions.
Position Vectors
Position vectors represent the location of a particle in a coordinate space, indicated by components in the x and y directions. Vectors show both magnitude and direction.

For a particle beginning motion at the origin with no initial speed along the x-axis and 10 m/s along the y-axis, the position vectors can be calculated using kinematic equations:
  • \(x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \)
  • \(y(t) = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \)
At \(t = 2\) seconds as previously calculated:
  • \(x(2) = \frac{1}{2} \cdot 8 \cdot 2^2 = 16 \text{ m}\)
  • \(y(2) = 10 \cdot 2 + \frac{1}{2} \cdot 2 \cdot 2^2 = 20 + 2 = 22 \text{ m}\)
Thus, the position vector of the particle at this time is \(\mathbf{r} (t) = 16 \hat{\mathbf{i}} + 22 \hat{\mathbf{j}} \), describing its precise location in the plane.

Understanding position vectors helps visualize and calculate where the particle is moving over time.

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Most popular questions from this chapter

Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

A passenger arriving in a new town wishes to go from the station to a hotel located \(10 \mathrm{~km}\) away on a straight road from the station. A dishonest cabman takes him along a circuitous path \(23 \mathrm{~km}\) long and reaches the hotel in \(28 \mathrm{~min}\). What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

The ceiling of a long hall is \(25 \mathrm{~m}\) high. What is the maximum horizontal distance that a ball thrown with a speed of \(40 \mathrm{~m} \mathrm{~s}^{-1}\) can go without hitting the ceiling of the hall?

Read each statement below carefully and state with reasons, if it is true or false: (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.

On an open ground, a motorist follows a track that turns to his left by an angle of \(60^{\circ}\) after every \(500 \mathrm{~m}\). Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
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