Problem 3 - Thomas Calculus

Question

Exercises \(1-6\) give the positions \(s=f(t)\) of a body moving on a coordinate line, with \(s\) in meters and \(t\) in seconds. a. Find the body's displacement and average velocity for the given time interval. b. Find the body's speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? $$ s=-t^{3}+3 t^{2}-3 t, \quad 0 \leq t \leq 3 $$

Step-by-Step Solution

Answer

Displacement: -9 m, Avg. velocity: -3 m/s, No direction change.

Step 1: Calculate Displacement

Displacement is the change in position over the time interval given by \( s(t_2) - s(t_1) \). Here, we need to find the position at \( t=3 \) and \( t=0 \).First, substitute \( t=3 \) into the function: \[s(3) = -(3)^3 + 3(3)^2 - 3(3) = -27 + 27 - 9 = -9 \]Now substitute \( t=0 \):\[s(0) = -(0)^3 + 3(0)^2 - 3(0) = 0 \]So, the displacement is: \[-9 - 0 = -9 \, \text{meters}.\]

Step 2: Calculate Average Velocity

Average velocity is the displacement divided by the time interval: \[v_{\text{avg}} = \frac{s(3) - s(0)}{3 - 0}\]Substitute the calculated displacement:\[v_{\text{avg}} = \frac{-9 - 0}{3 - 0} = \frac{-9}{3} = -3 \, \text{m/s}.\]

Step 3: Calculate Velocity (Speed) and Acceleration

To find speed and acceleration at the endpoints, we first need the velocity \( v(t) = s'(t) \) and acceleration \( a(t) = v'(t) \).First, calculate the derivative of \( s(t) \):\[v(t) = \frac{d}{dt}(-t^3 + 3t^2 - 3t) = -3t^2 + 6t - 3\]Then calculate the velocity at \( t=0 \) and \( t=3 \):\[v(0) = -3(0)^2 + 6(0) - 3 = -3 \, \text{m/s},\]\[v(3) = -3(3)^2 + 6(3) - 3 = -27 + 18 - 3 = -12 \, \text{m/s}.\]Acceleration is the derivative of velocity:\[a(t) = \frac{d}{dt}(-3t^2 + 6t - 3) = -6t + 6\]Calculate acceleration at \( t=0 \) and \( t=3 \):\[a(0) = -6(0) + 6 = 6 \, \text{m/s}^2,\]\[a(3) = -6(3) + 6 = -18 + 6 = -12 \, \text{m/s}^2.\]

Step 4: Determine Direction Change

The body changes direction when velocity changes its sign. We need to find when \( v(t) = 0 \).Set \( v(t) = 0 \) and solve for \( t \):\[-3t^2 + 6t - 3 = 0 \]This simplifies to:\[(t - 1)^2 = 0 \]So, \( t = 1 \).Check if the sign of \( v(t) \) changes around \( t=1 \): - For \( t < 1 \), e.g., \( t = 0 \), \( v(0) = -3 \) (negative). - For \( t > 1 \), e.g., \( t = 2 \), \( v(2) = -3(2)^2 + 6(2) - 3 = -12 + 12 - 3 = -3 \) (still negative).The velocity does not change sign, hence the body doesn't change direction in \([0,3]\).

Key Concepts

Displacement

In calculus, displacement refers to the change in position of a body along a straight line over a certain time interval. It is calculated as the difference between the final and initial positions. For this specific problem, we have a function representing the position of the body: \[ s(t) = -t^3 + 3t^2 - 3t \] - To find the displacement from \( t = 0 \) to \( t = 3 \), we evaluate the function at these points: - \( s(3) = -(3)^3 + 3(3)^2 - 3(3) = -9 \) - \( s(0) = 0 \)- Thus, the displacement is: - \( s(3) - s(0) = -9 - 0 = -9 \text{ meters} \)
- Displacement, therefore, is not just the distance travelled. It's the net change considering the starting and stopping point in a straight path.

Average Velocity

Average velocity gives us a measure of how fast the body is moving, considering the total displacement over a given time interval. It is the ratio of displacement to the time interval duration.- From the problem, we have already calculated the displacement to be \(-9\) meters over the time interval \(0 \leq t \leq 3\).- The formula for average velocity is: \[ v_{\text{avg}} = \frac{s(3) - s(0)}{3 - 0} \] - Plug in the values: \[ v_{\text{avg}} = \frac{-9 - 0}{3} = -3 \text{ m/s} \]This means that overall, the body moved at an average rate of \(-3 \text{ meters per second}\), with the negative sign indicating the direction of motion was backwards relative to the coordinate line.

Acceleration

Acceleration is the rate of change of velocity with respect to time. In calculus, this is represented by the derivative of the velocity function.- First, we determine the velocity function by taking the derivative of the position function \(s(t)\): \[ v(t) = \frac{d}{dt}(-t^3 + 3t^2 - 3t) = -3t^2 + 6t - 3 \] - Next, the acceleration function is found by differentiating the velocity function: \[ a(t) = \frac{d}{dt}(-3t^2 + 6t - 3) = -6t + 6 \] - To find the acceleration at the endpoints \(t = 0\) and \(t = 3\): - \( a(0) = -6(0) + 6 = 6 \text{ m/s}^2 \) - \( a(3) = -6(3) + 6 = -12 \text{ m/s}^2 \)These values indicate:- At \( t = 0 \), the body is accelerating positively.- At \( t = 3 \), the body is decelerating or accelerating negatively. This change in acceleration could impact how the body's velocity changes over time.