Questions: The function p(t)=t^3-15t^2+72t gives the position, relative to its starting point, of an object moving along a straight line. Identify the time intervals when the object is moving in the positive and the negative direction. Combine multiple intervals with the union symbol.

The function p(t)=t^3-15t^2+72t gives the position, relative to its starting point, of an object moving along a straight line. Identify the time intervals when the object is moving in the positive and the negative direction. Combine multiple intervals with the union symbol.

Transcript text: The function $p(t)=t^{3}-15 t^{2}+72 t$ gives the position, relative to its starting point, of an object moving along a straight line. Identify the time intervals when the object is moving in the positive and the negative direction. Combine multiple intervals with the union symbol.

Solution

Solution Steps

Step 1: Find the velocity function

The velocity $ v(t) $ is the derivative of the position function $ p(t) $. Compute $ v(t) $ as follows: \[ v(t) = \frac{d}{dt} \left( t^{3} - 15t^{2} + 72t \right) = 3t^{2} - 30t + 72. \]

Step 2: Determine when the velocity is zero

Set $ v(t) = 0 $ and solve for $ t $: \[ 3t^{2} - 30t + 72 = 0. \] Divide the equation by 3: \[ t^{2} - 10t + 24 = 0. \] Factor the quadratic equation: \[ (t - 4)(t - 6) = 0. \] Thus, $ t = 4 $ and $ t = 6 $ are the critical points.

Step 3: Test intervals to determine the sign of the velocity

The critical points divide the time axis into three intervals: $ (-\infty, 4) $, $ (4, 6) $, and $ (6, \infty) $. Test the sign of $ v(t) $ in each interval:

For $ t \in (-\infty, 4) $, choose $ t = 0 $: \[ v(0) = 3(0)^{2} - 30(0) + 72 = 72 > 0. \] The object is moving in the positive direction.
For $ t \in (4, 6) $, choose $ t = 5 $: \[ v(5) = 3(5)^{2} - 30(5) + 72 = 75 - 150 + 72 = -3 < 0. \] The object is moving in the negative direction.
For $ t \in (6, \infty) $, choose $ t = 7 $: \[ v(7) = 3(7)^{2} - 30(7) + 72 = 147 - 210 + 72 = 9 > 0. \] The object is moving in the positive direction.

Step 4: Combine intervals

The object is moving in the positive direction for $ t \in (-\infty, 4) \cup (6, \infty) $ and in the negative direction for $ t \in (4, 6) $.