Questions: An object moves along a horizontal coordinate line in such a way that its position at time t is specified by s=t^3-12t^2+45t+9. Here s is measured in feet and t in seconds. (a) When is the velocity 0 ? t= 3 5 or t> (b) When is the velocity positive? t< 3 <t< (c) When is the object moving to the left (that is, in the negative direction)? 5 (d) When is the acceleration positive? t> 3

Solution Steps

The velocity \( v(t) \) is the derivative of the position function \( s(t) \). Given \( s(t) = t^3 - 12t^2 + 45t + 9 \), we find the derivative:

\[ v(t) = \frac{d}{dt}(t^3 - 12t^2 + 45t + 9) = 3t^2 - 24t + 45 \]

To find when the velocity is zero, we solve \( v(t) = 0 \):

\[ 3t^2 - 24t + 45 = 0 \]

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ t = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 3 \cdot 45}}{2 \cdot 3} = \frac{24 \pm \sqrt{576 - 540}}{6} = \frac{24 \pm \sqrt{36}}{6} = \frac{24 \pm 6}{6} \]

This gives us:

\[ t = \frac{30}{6} = 5 \quad \text{and} \quad t = \frac{18}{6} = 3 \]

So, the velocity is zero at \( t = 3 \) and \( t = 5 \).

\[ \boxed{t = 3 \text{ or } t = 5} \]

To determine when the velocity is positive, we analyze the sign of \( v(t) = 3t^2 - 24t + 45 \) around the critical points \( t = 3 \) and \( t = 5 \).

• For \( t < 3 \), choose \( t = 0 \): \[ v(0) = 3(0)^2 - 24(0) + 45 = 45 \quad (\text{positive}) \]

• For \( 3 < t < 5 \), choose \( t = 4 \): \[ v(4) = 3(4)^2 - 24(4) + 45 = 48 - 96 + 45 = -3 \quad (\text{negative}) \]

• For \( t > 5 \), choose \( t = 6 \): \[ v(6) = 3(6)^2 - 24(6) + 45 = 108 - 144 + 45 = 9 \quad (\text{positive}) \]

Thus, the velocity is positive for \( t < 3 \) and \( t > 5 \).

\[ \boxed{t < 3 \text{ or } t > 5} \]

The object is moving to the left when the velocity is negative. From the previous analysis, we found that \( v(t) \) is negative for \( 3 < t < 5 \).

\[ \boxed{3 < t < 5} \]

Final Answer