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Chapter 3: Problem 87

A particle has a constant acceleration of \(5.0 \mathrm{m} / \mathrm{s}^{2}\) to the east. At time \(t=0,\) it is \(2.0 \mathrm{m}\) east of the origin and its velocity is \(20 \mathrm{m} / \mathrm{s}\) north. What are the components of its position vector at \(t=2.0 \mathrm{s} ?\)

Short Answer

Expert verified
Answer: The components of the position vector at t=2 seconds are 12.0 m eastward and 40.0 m northward.

Step by step solution

01

Determine the eastward component of position

Since the particle's motion in the eastward direction is under constant acceleration, we can use the equation: \(x(t) = x_0 + v_{0x}t + \frac{1}{2}at^2\). Here, \(x_0 = 2.0 m\), \(v_{0x} = 0\) (Because the initial velocity is in the northward direction, there is no eastward component of initial velocity), and \(a = 5.0 \mathrm{m} / \mathrm{s}^2\). Now we can plug in these values and given time, \(t=2s\), into the equation to get the eastward component of the position: \(x(2) = 2.0 + 0(2) + \frac{1}{2}(5)(2)^2 = 2.0 + 10 = 12.0 \mathrm{m}\) eastward.
02

Determine the northward component of position

In the northward direction, since there is no acceleration, the particle moves with constant velocity. We can use the equation: \(y(t) = y_0 + v_{0y}t\) Here, \(y_0 = 0\) (particle starts from the origin, so there is no initial position in the northward direction) and \(v_{0y} = 20 \mathrm{m} / \mathrm{s}\). Now we can plug in these values and given time, \(t=2s\), into the equation to get the northward component of the position: \(y(2) = 0 + 20(2) = 40 \mathrm{m}\) northward.
03

Write the position vector components

After calculating both eastward and northward components of position, we have the position vector at \(t=2.0s\): \((\)12.0 m eastward, 40.0 m northward\()\).

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