Suppose an object moves along a line with velocity (in \( \mathrm{ft} / \mathrm{s} \) ) \( \mathrm{v}(\mathrm{t})=6-4 \mathrm{t} \), for \( 0 \leq \mathrm{t} \leq 3 \), where t is measured in seconds. The motion is in the negative direction on the interval \( (1.5,3] \). (Type your answer in interval notation.) b. The displacement on \( [0,3] \) is \( \square \) feet.

To find the displacement of the object over the interval \([0,3]\), we need to integrate the velocity function \(v(t) = 6 - 4t\) with respect to time \(t\) from \(0\) to \(3\): \[ \text{Displacement} = \int_0^3 v(t) \, dt = \int_0^3 (6 - 4t) \, dt. \] Calculating the integral: \[ \int (6 - 4t) \, dt = 6t - 2t^2 + C. \] Now we evaluate this from \(0\) to \(3\): \[ \left[6(3) - 2(3)^2\right] - \left[6(0) - 2(0)^2\right] = (18 - 18) - (0) = 0. \] Thus, the displacement on \([0,3]\) is \(0\) feet. Additionally, it's interesting to note that the object returns to its original position after the motion, interpreting the interplay between positive and negative velocity in this time span. Moving through scenarios where an object moves back and forth can often lead to a zero net displacement, which is fun to explore in physics—they're like a dance of motion! Another fun fact: in real-world applications, understanding displacement versus distance traveled is crucial! Take a road trip where you drive in circles; you might cover many miles, but your displacement is just the straight-line distance from your starting point to your home!