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Chapter 1: Problem 7

If the position function of a particle is \(x(t)=2 t^{3}-6 t^{2}+12 t-18, t>0,\) find when the particle is changing direction.

Short Answer

Expert verified
The particle changes direction at \(t = 1\) and \(t = 2\).

Step by step solution

01

Compute the velocity

The velocity function is the first derivative of the position function. Hence, differentiate the position function: \(v(t) = x'(t) = 6t^2 - 12t + 12\). This function gives the velocity of the particle at any point in time.
02

Find possible points of direction change

The direction changes when the velocity changes signs, which occurs when the velocity is zero. Hence, find when the velocity is zero: \(6t^2 - 12t + 12 = 0\). Factor this equation fully to solve for \(t\). After factoring and simplifying, two possible values for \(t\) are obtained: \(t = 1\), and \(t = 2\). These are the points in time where the direction of the particle might change.
03

Test the intervals

Now, verify the intervals before, between and after the roots obtained: \(t < 1\), \(1 < t < 2\), and \(t > 2.\) Substitute values from these intervals into the velocity function. For interval \(t < 1\), \(t = 0\) could be used, for interval \(1 < t < 2\), \(t = 1.5\) could be used and for \(t > 2\), \(t = 3\) could be used. Velocity at \(t = 0\) and \(t = 3\) is positive and at \(t = 1.5\) is negative, which means the sign changes and hence the particle changes direction at \(t = 1\) and \(t = 2\).

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