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Chapter 12: Problem 74

A particle, originally at rest and located at point \((3 \mathrm{ft}\) \(2 \mathrm{ft}, 5 \mathrm{ft}),\) is subjected to an acceleration \(\mathbf{a}=\left\\{6 t \mathbf{i}+12 t^{2} \mathbf{k}\right\\} \mathrm{ft} / \mathrm{s}^{2}\) Determine the particle's position \((x, y, z)\) when \(t=2 \mathrm{s}\).

Short Answer

Expert verified
The position of the particle at the specified time (t = 2s) is (11 ft, 2 ft, 21 ft).

Step by step solution

01

Noting down the Initial Condition

From the problem, the particle is initially at rest. This means the initial velocity \( \mathbf{v}_{0} = 0 \), at \( t=0 \). The initial position is \( (x_0, y_0, z_0) = (3ft, 2ft, 5ft) \).
02

Express Acceleration in Component Form

The acceleration given is \( \mathbf{a}=6t \mathbf{i}+12t^{2} \mathbf{k} \) units. In component form, acceleration can be expressed as \( \mathbf{a}(t)=(a_x(t), a_y(t), a_z(t))=(6t, 0, 12t^2) \) ft/s\(^2\).
03

Calculate Velocity by Integrating Acceleration

Velocity is obtained by the integral of acceleration. Therefore, integrate the acceleration \( \mathbf{a}(t) \). For each component, x, y, and z, \( v_x(t)=\int a_x(t) dt = \int 6t dt = 3t^2+C_1 \), \( v_y(t)=0 \) since \( a_y(t)=0 \), \( v_z(t)=\int a_z(t) dt = \int 12t^2 dt = 4t^3 + C_2 \). Since the particle was initially at rest at \( t = 0 \), \( C_1 = 0 \) and \( C_2 = 0 \). Hence, the velocity \( \mathbf{v}(t) \) becomes \( \mathbf{v}(t)=(3t^2, 0, 4t^3) \) ft/s.
04

Calculate Position by Integrating Velocity

Position is obtained by integrating velocity. Therefore, integrate the velocity \( \mathbf{v}(t) \). For each component, \( x(t)=\int v_x(t) dt = \int 3t^2 dt = t^3 + x_0 \), \( y(t) = 2 \) ft (since \( v_y=0 \)), \( z(t)=\int v_z(t) dt = \int 4t^3 dt = t^4 + z_0 \). Substitute the initial position values \( x_0 = 3 \) ft, \( y_0 = 2 \) ft, and \( z_0 = 5 \) ft to result in \( \mathbf{r}(t) = (t^3 + 3, 2, t^4 + 5) \) ft.
05

Determine Position at t = 2 seconds

Substitute \( t = 2 \) s into the position function \( \mathbf{r}(t)\): \( \mathbf{r}(2) = ((2)^3+3, 2, (2)^4+5) = (11, 2, 21) \) ft

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Key Concepts

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

Position Calculation
To understand the position calculation of a particle, we must follow the changes the particle undergoes over time. Initially, the particle starts at a specific position, in this case,
  • Component positions: \( x_0 = 3 \text{ ft} \), \( y_0 = 2 \text{ ft} \), \( z_0 = 5 \text{ ft} \)
These are essential to track any future movements.
As time progresses and with given initial conditions, we integrate velocity to find the position functions for each component. Here, the calculations transform
  • \( x(t) = t^3 + 3 \)
  • \( y(t) = 2 \) (a constant, since there's no movement in the \( y \)-direction)
  • \( z(t) = t^4 + 5 \)
By substituting \( t = 2 \text{ seconds} \), we find the particle's position at that time:
  • \( x = 11 \text{ ft} \)
  • \( y = 2 \text{ ft} \)
  • \( z = 21 \text{ ft} \)
Thus, the particle is positioned at \((11, 2, 21)\) ft when \( t = 2 \text{ s} \). The position change reflects how the acceleration and velocities have influenced the particle's journey.
Acceleration Integration
Integration of acceleration is a fundamental step in determining a particle's motion, involving the calculation of velocity over time.Given the acceleration components
  • \( a_x(t) = 6t \)
  • \( a_y(t) = 0 \)
  • \( a_z(t) = 12t^2 \)
We integrate each component to obtain the velocity:
  • \( v_x(t) = \int 6t \, dt = 3t^2 + C_1 \)
  • \( v_y(t) = 0 \quad ( \text{since}\, a_y(t) = 0) \)
  • \( v_z(t) = \int 12t^2 \, dt = 4t^3 + C_2 \)
Integration serves as a bridge, transforming acceleration into velocity by considering the rate of change over time. Through understanding this process, one can then proceed to deduce the particle's future path or position.
Velocity Determination
Velocity determination involves the conversion of raw acceleration data into actionable velocity values, describing the particle's path.When we integrate acceleration:
  • \( v_x(t) = 3t^2 + C_1 \)
  • \( v_y(t) = 0 \)
  • \( v_z(t) = 4t^3 + C_2 \)
Initial conditions are crucial in computing the constants of integration. Since the particle is initially at rest, both \( C_1 \) and \( C_2 \) resolve to zero:
  • \( v_x(t) = 3t^2 \)
  • \( v_z(t) = 4t^3 \)
These equations outline how velocity can change with time and integrate into the broader mechanics of motion.
Initial Conditions in Mechanics
Initial conditions in mechanics establish a baseline from which motion unfolds, such as initial velocity and position.For our particle:
  • Initial position: \( (3, 2, 5) \) ft
  • Initial velocity: \( \mathbf{v}_0 = (0, 0, 0) \) since it was at rest
These elements define how the calculations proceed. With no initial velocity, the constants of integration for velocity (\( C_1 \) and \( C_2 \)) become zero, simplifying further calculations.
Initial conditions also guide the determination of
  • Position: \( \mathbf{r}(t) = (t^3 + 3, 2, t^4 + 5) \)
Therefore, understanding these initial conditions is key to predicting future movement and calculating precise trajectories.

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Most popular questions from this chapter

For a short time the arm of the robot is extending such that \(\quad \dot{r}=1.5 \mathrm{ft} / \mathrm{s}\) when \(r=3 \mathrm{ft}, z=\left(4 t^{2}\right) \mathrm{ft},\) and \(\theta=0.5 t \mathrm{rad},\) where \(t\) is in seconds. Determine the magnitudes of the velocity and acceleration of the grip \(A\) when \(t=3 \mathrm{s}\).

A man riding upward in a freight elevator accidentally drops a package off the elevator when it is \(100 \mathrm{ft}\) from the ground. If the elevator maintains a constant upward speed of \(4 \mathrm{ft} / \mathrm{s}\), determine how high the elevator is from the ground the instant the package hits the ground. Draw the \(v\) -t curve for the package during the time it is in motion. Assume that the package was released with the same upward speed as the elevator.

The velocity of a particle traveling along a straight line is \(v=v_{0}-k s,\) where \(k\) is constant. If \(s=0\) when \(t=0\) determine the position and acceleration of the particle as a function of time.

An airplane is flying in a straight line with a velocity of \(200 \mathrm{mi} / \mathrm{h}\) and an acceleration of \(3 \mathrm{mi} / \mathrm{h}^{2} .\) If the propeller has a diameter of 6 ft and is rotating at a constant angular rate of 120 rad/s, determine the magnitudes of velocity and acceleration of a particle located on the tip of the propeller.

A man walks at \(5 \mathrm{km} / \mathrm{h}\) in the direction of a \(20 \mathrm{km} / \mathrm{h}\) wind. If raindrops fall vertically at \(7 \mathrm{km} / \mathrm{h}\) in still air, determine direction in which the drops appear to fall with respect to the man.
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