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Chapter 4: Problem 4

A particle leaves the origin at \(t=0\) with an initial velocity \(\overrightarrow{\mathbf{v}}_{0}=(3.6 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}\). It experiences a constant acceleration \(\overrightarrow{\mathbf{a}}=-\left(1.2 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathbf{i}}-\left(1.4 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathbf{j}} .(a)\) At what time does the particle reach its maximum \(x\) coordinate? \((b)\) What is the velocity of the particle at this time? \((c)\) Where is the particle at this time?

Short Answer

Expert verified
The particle reaches its maximum x-coordinate at \(t = 3\,s\). The velocity of the particle at this point is \(0\,\hat{i} - 4.2\,\hat{j}\, m/s\), and the position of the particle at this time is \(0\,\hat{i} - 6.3\,\hat{j}\,m\)

Step by step solution

01

Determine time at maximum x-coordinate (Solution to part (a))

The x-component of motion can be solved independently from the y-component as the two are perpendicular to each other. The equation for displacement in the x-direction under constant acceleration is \(x = v_0t + 0.5at^2\). Here, \(v_0\) is the x-component of the initial velocity, \(a\) is the x-component of the acceleration, and \(t\) is the time. To find the time at which the particle reaches its maximum x coordinate, we can set the derivative of the displacement function with respect to time equal to zero. Since \(x' = v_0 + at\) , when \(x'\) equals zero, we get \(t = -v_0/a\). Substituting \(v_0 = 3.6\,m/s\) and \(a = -1.2\,m/s^2\), we get \(t = 3\,s\).
02

Determine velocity at maximum x-coordinate (Solution to part (b))

The velocity in the x-direction at any time, \(t\), under constant acceleration can be found using the equation \(v = v_0 + at\). Substituting for \(v_0 = 3.6\,m/s\), \(a = -1.2\,m/s^2\) and \(t = 3\,s\) yields \(v_x = 0\,m/s\). The velocity in the y-direction can be found using the same equation, with \(v_0 = 0\), \(a = -1.4\, m/s^2\) and \(t = 3\,s\), providing \(v_y = -4.2\,m/s\). Thus, the velocity of the particle when it reaches max x-coordinate is \((0, -4.2)\,m/s\) or \(0\,\hat{i} - 4.2\,\hat{j}\,m/s\).
03

Determine the position at maximum x-coordinate (Solution to part (c))

We need to find the y-coordinate at \(t = 3\,s\). For the y component of the motion, \(v_0 = 0\) and \(a = -1.4\,m/s^2\). Using the equation \(y = v_0t + 0.5at^2\), and substituting for \(t = 3\,s\), \(v_0 = 0\), and \(a = -1.4\,m/s^2\) gives \(y = -6.3\,m\). Given that the maximum x-coordinate is obtained when x = 0, the particle's position at \(t = 3\,s\) is \((0, -6.3)\,m\) or \(0\,\hat{i} - 6.3\,\hat{j}\,m\).

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