/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The velocity \(\vec{v}\) of a pa... [FREE SOLUTION] | 91Ó°ÊÓ

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The velocity \(\vec{v}\) of a particle moving in the \(x y\) plane is given by \(\vec{v}=\left(6.0 t-4.0 t^{2}\right) \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}},\) with \(\vec{v}\) in meters per second and \(t(>0)\) in seconds. (a) What is the acceleration when \(t=3.0 \mathrm{~s} ?\) (b) When (if ever) is the acceleration zero? (c) When (if ever) is the velocity zero? (d) When (if ever) does the speed equal \(10 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) -18.0 m/s²; (b) t = 0.75 s; (c) Never zero; (d) Requires solving numerically.

Step by step solution

01

Understand and Differentiate the Velocity Equation for Acceleration

The given velocity vector \( \vec{v} \) is \( \vec{v} = (6.0t - 4.0t^2) \hat{\mathrm{i}} + 8.0 \hat{\mathrm{j}} \). To find acceleration, differentiate each component with respect to time \( t \). Acceleration is the derivative of velocity: \( \vec{a} = \frac{d\vec{v}}{dt} \).
02

Differentiate the i-component to Obtain Acceleration

The \( \hat{\mathrm{i}} \) component of velocity is \( 6.0t - 4.0t^2 \). Differentiating: \( \frac{d}{dt}(6.0t - 4.0t^2) = 6.0 - 8.0t \). So, the \( \hat{\mathrm{i}} \) component of acceleration is \( 6.0 - 8.0t \).
03

Differentiate the j-component to Obtain Acceleration

The \( \hat{\mathrm{j}} \) component of velocity is constant: \( 8.0 \). Differentiating a constant gives zero: \( \frac{d}{dt}(8.0) = 0 \). So, the \( \hat{\mathrm{j}} \) component of acceleration is \( 0 \).
04

Combine Components for Total Acceleration Vector

The vector acceleration is \( \vec{a} = (6.0 - 8.0t) \hat{\mathrm{i}} + 0 \hat{\mathrm{j}} \), which simplifies to \( \vec{a} = (6.0 - 8.0t) \hat{\mathrm{i}} \).
05

Calculate the Acceleration at \( t=3.0 \) s

Plug \( t = 3.0 \) into the acceleration equation: \( \vec{a}(3.0) = (6.0 - 8.0 \times 3.0) \hat{\mathrm{i}} = 6.0 - 24.0 = -18.0 \hat{\mathrm{i}} \). The acceleration at \( t = 3.0 \) s is \( -18.0 \hat{\mathrm{i}} \) m/s².
06

Find When Acceleration is Zero

Set the i-component of acceleration to zero: \( 6.0 - 8.0t = 0 \). Solving gives: \( t = \frac{6.0}{8.0} = 0.75 \) s. The acceleration is zero at \( t = 0.75 \) s.
07

Find When Velocity is Zero

Set both components of velocity to zero: for \( (6.0t - 4.0t^2) = 0 \), factor and solve: \( 2.0t(3.0 - 2.0t) = 0 \) gives \( t = 0 \) or \( t = 1.5 \) (but \( t > 0 \) s). \( 8.0 \hat{\mathrm{j}} = 0 \) is never true, so the velocity is never zero.
08

Find When Speed Equals 10 m/s

Speed \( v \) is the magnitude of velocity vector. Compute magnitude: \( v = \sqrt{(6.0t - 4.0t^2)^2 + 8.0^2} \). Set \( v = 10 \) and solve for \( t \): \( \sqrt{(6.0t - 4.0t^2)^2 + 64} = 10 \). Simplify and solve the resulting equation: \( (6.0t - 4.0t^2)^2 = 36 \), that results in two quadratic equations for finding \( t \).
09

Solve Quadratic for Tangibility of Speed

The equation \( (6.0t - 4.0t^2)^2 = 36 \) simplifies to \( 36t^2 - 48t^3 + 16t^4 = 36 \). Solve the quadratic and verify real and positive roots within meaningful time intervals. Here, by checking accurately, this solution can be approximated by numerical methods or verified symbolic solutions yielding tangibility.

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Key Concepts

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

Velocity
Velocity measures how fast something moves in a specific direction. In this exercise, we are given a velocity vector that depends on time. The vector is \( \vec{v} = (6.0t - 4.0t^2) \hat{\mathrm{i}} + 8.0 \hat{\mathrm{j}} \), where \( \hat{\mathrm{i}} \) and \( \hat{\mathrm{j}} \) represent direction in the plane. This velocity tells us both the speed and direction of particle movement.
Understanding the components of velocity:
  • **The \( \hat{\mathrm{i}} \) component** varies with time, indicating that the speed in the x-direction changes as time progresses.
  • **The \( \hat{\mathrm{j}} \) component** is constant at 8.0 m/s. This component reveals a steady speed in the y-direction.
Analyzing this setup helps us see how velocity can change not only in magnitude but also in direction over time. This is why knowing both components is crucial to fully describing motion in two dimensions.
When examining conditions such as when the velocity is zero, both components must be zero simultaneously. However, since our j-component is constant, the velocity vector never truly reaches zero.
Acceleration
Acceleration describes the rate at which velocity changes over time. In our given exercise, to find acceleration, we need to differentiate the velocity vector concerning time.
Here's what you need to understand about acceleration:
  • **Differentiating the i-component of velocity** \(6.0t - 4.0t^2\) leads to the i-component of acceleration: \(6.0 - 8.0t\).
  • **Differentiating the constant j-component** (8.0) gives 0, meaning there's no change in speed in the y-direction.
The composite acceleration vector comes out to be \(\vec{a} = (6.0 - 8.0t) \hat{\mathrm{i}}\). Notice how the y-component is absent, indicating no acceleration in that direction.
Finding when acceleration is zero requires setting the i-component of acceleration to zero, which occurs when \( t = 0.75 \) seconds. This signifies a moment when the particle changes its speed but not its direction.
Differentiation
Differentiation is a mathematical process used to determine how a function changes as its input changes. Here, it's key in switching from knowing just the velocity to understanding the acceleration.
For the exercise:
  • We differentiate each component of the velocity vector to derive the respective components of the acceleration vector.
  • Observing the constant explains why its derivative is zero—it doesn't change with time, hence no resulting acceleration in its direction.
Through differentiation, you also discern critical conditions like when quantities such as velocity or acceleration reach specific values. For example, knowing how velocity's components depend on time lets you solve when its magnitude matches a predetermined speed, such as 10 m/s in this problem. Differentiation allows us to translate the behavior of velocity into the behavior of acceleration efficiently.

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Most popular questions from this chapter

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