Question

The velocity of a particle moving in the \(x y\) plane is given by \(\overrightarrow{\mathbf{v}}=\left[\left(6.0 \mathrm{~m} / \mathrm{s}^{2}\right) t-\left(4.0 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\right] \hat{\mathbf{i}}+(8.0 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\). Assume \(t>0 .(a)\) What is the acceleration when \(t=3 \mathrm{~s} ?(b)\) 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

\nThe acceleration when \(t=3\,\mathrm{s}\) is \( -18\hat{i}\,\mathrm{m/s^{2}}\),\nThe acceleration is zero at \(t = 0.75s\),\nThe particle stops moving in the \(x\)-direction at \(t = 1.5s\),\nThe particle's speed is \(10\,\mathrm{m/s}\) at \(t = 0.69s\) and again at \(t = 2.08s\)

Step-by-step solution

Find the Acceleration Vector

The acceleration vector \(\overrightarrow{a}\) is the derivative of the velocity vector \(\overrightarrow{v}\). So, \(\overrightarrow{a} = \frac {d\overrightarrow{v}} {dt}\).\nThe derivative of the velocity expression \(\overrightarrow{v} =\left[\left(6.0 \mathrm{~m} /s^{2}\right) t -\left(4.0 \mathrm{~m} / s^{3}\right)t^{2}\right] \hat{i} +(8.0 \mathrm{~m} / s) \hat{j}\) leads to acceleration vector \(\overrightarrow{a} = [6 - 8t] \hat{i}\)

Compute Acceleration at \(t=3s\)

To find the acceleration when \(t=3s\), substitute \(t = 3\) in acceleration expression. \(\overrightarrow{a}(3) = [6 - 8*3] \hat{i} = -18 \hat{i}\)

Find When Acceleration is Zero

To find when the acceleration is zero, equate the acceleration expression to zero and solve for \(t\), giving \([6 - 8t] = 0 => t = 0.75s\)

Find When Velocity is Zero

To find when the velocity is zero, we need to set each of the components of the velocity equation to zero. The equation \(\left(6t-4t^{2}\right)=0\) gives \(t = 0s\) and \(t = 1.5s\). However, since \(t>0\), the only valid solution is \(t = 1.5s\). The \(y\)-component of velocity \(8.0 m/s\) is constant and hence, never equals zero.

Find When Speed is 10m/s

The speed is the magnitude of the velocity vector, it is obtained by square-rooting the sum of squares of each term in the velocity expression. Mathematically, \(|v| = \sqrt{\left(6.0m/s^{2}t - 4.0m/s^{3}t^{2}\right)^{2} + (8.0m/s)^{2}}\). Equating \(|v|\) to \(10m/s\) and solving this quadratic equation gives \(t = 0.69s\) and \(t = 2.08s\)

Key concepts

Velocity Vector

In physics, a vector is a quantity with both a magnitude and a direction. The velocity vector of a particle conveys information about how fast and in which direction the particle is moving.

For a particle moving in a plane, the velocity vector is typically expressed in terms of its horizontal (\( \, \hat{i} \)) and vertical (\( \, \hat{j} \)) components:

\[\overrightarrow{\mathbf{v}} = v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}\]

In the given exercise, the velocity vector is:

• \( [6.0 \, m/s^2 \cdot t - 4.0 \, m/s^3 \cdot t^2] \hat{\mathbf{i}} + 8.0 \, m/s \hat{\mathbf{j}} \).

This means the horizontal velocity changes with time due to the \( t^2 \) term, while the vertical component is constant. To solve problems about velocity, always pay attention to both components, especially their dependence on time.

Acceleration Vector

The acceleration vector represents the rate at which the velocity of a particle changes over time. Acceleration is also a vector quantity with both magnitude and direction.

To calculate the acceleration vector, we need to take the derivative of the velocity vector with respect to time (\( t \)):

• \[\overrightarrow{a} = \frac {d\overrightarrow{v}} {dt}\]

This implies differentiating each component of the velocity vector separately. For the exercise given, the velocity vector differentiates to yield:

• \[ \overrightarrow{a} = [6 - 8t] \hat{\mathbf{i}} \]

The constant vertical velocity component means that there is no acceleration in the \( \hat{\mathbf{j}} \) direction.

This simplified form of the acceleration vector shows that only the horizontal component changes over time, and we can calculate it at specific times by substituting values of \( t \).

Speed Calculation

Speed represents the magnitude of the velocity vector and conveys how fast an object is moving regardless of direction. Unlike velocity, speed is a scalar, meaning it has only a magnitude, not direction.

To calculate speed from a velocity vector, use the Pythagorean theorem to find the magnitude:

• \[|v| = \sqrt{v_x^2 + v_y^2}\]

Substituting the given components:

• 
  \[|v| = \sqrt{(6.0 \, m/s^2 \cdot t - 4.0 \, m/s^3 \cdot t^2)^2 + (8.0 \, m/s)^2} \].

This expression provides a method to find the speed at any given time. For example, to find when the speed equals a specific value, like 10 \( m/s \), substitute this into the expression and solve for \( t \), often yielding multiple possible times.

Derivative of Velocity

The derivative of a function describes how the function changes with respect to one of its variables—in this case, time (\( t \)).

The derivative of the velocity vector gives us the acceleration vector, indicating how quickly the velocity is changing. Using the derivative is crucial in finding the acceleration when dealing with time-dependent velocity expressions.

For example, differentiating the velocity vector:

• \[\overrightarrow{v} = (6.0 \, m/s^2 \cdot t - 4.0 \, m/s^3 \cdot t^2) \hat{\mathbf{i}} + (8.0 \, m/s) \hat{\mathbf{j}}\]

With respect to time gives:

• \[\overrightarrow{a} = (6 - 8t) \hat{\mathbf{i}}\]

This illustrates the process of taking a derivative and helps understand how changes in velocity affect the motion of the particle over time.