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Chapter 12: Problem 5

The velocity of a particle traveling in a straight line is given by \(v=\left(6 t-3 t^{2}\right) \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds. If \(s=0\) when \(t=0\), determine the particle's deceleration and position when \(t=3 \mathrm{~s}\). How far has the particle traveled during the 3-s time interval, and what is its average speed?

Short Answer

Expert verified
The deceleration of the particle at t=3s is -12 m/s^2. The position of the particle at t=3s is 0 m. During the 3-s time interval, the particle has not moved. Thus, its average speed is 0 m/s.

Step by step solution

01

Find the Position Function

The position function s(t) is the integral of the velocity function v(t). \[ s(t) = \int v(t) dt = \int (6t - 3t^2) dt. \] Apply the power rule to integrate: \[ s(t) = 3t^2 - t^3 + C, \] where C is the constant of integration. Given that s=0 when t=0, we find that C=0. So the position function is: \[ s(t) = 3t^2 - t^3. \]
02

Find the Deceleration at Time t = 3 seconds

The deceleration of the particle is the negative of the second derivative of the position function, or the derivative of the velocity function: \[ a(t) = -\frac{d^2s}{dt^2} = -\frac{dv}{dt}. \] Differentiate the velocity function to get: \[ a(t) = -\frac{d}{dt}(6t - 3t^2) = 6 - 6t. \] The deceleration at t = 3s is: \[ a(3) = 6 - 6(3) = -12 m/s^2. \]
03

Find the Position at Time t = 3 seconds

Substitute t = 3s into the position function: \[ s(3) = 3(3)^2 - (3)^3 = 27 - 27 = 0 \, m. \]
04

Find Total Distance Traveled

The total distance traveled by the particle during the 3 sec interval is the absolute value of the change in position. In this case: \[ s(3) - s(0) = 0 - 0 = 0 \, m. \]
05

Find Average Speed

The average speed is the total distance traveled divided by the time taken. Here: \[ \text{Average speed} = \frac{ \text{Total distance}}{ \text{Total time}} = \frac{0}{3} = 0 \, m/s. \]

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