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Chapter 2: Problem 19

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

Expert verified
\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

01

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 \(\overrightarrow{a} = [6 - 8t] \hat{i}\)
02

Compute Acceleration at \(t=3s\)

To find the acceleration when \(t=3s\), substitute \(t = 3\) in acceleration expression. \(\overrightarrow{a}(3) = [6 - 8*3] \hat{i} = -18 \hat{i}\)
03

Find When Acceleration is Zero

To find when the acceleration is zero, equate the acceleration expression to zero and solve for \(t\), giving \([6 - 8t] = 0 => t = 0.75s\)
04

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.
05

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 \(t = 2.08s\)

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