Problem 144 - Calculus Volume 1

Question

A car driving along a freeway with traffic has traveled \(s(t)=t^{3}-6 t^{2}+9 t\) meters in \(t\) seconds. a. Determine the time in seconds when the velocity of the car is 0. b. Determine the acceleration of the car when the velocity is 0.

Step-by-Step Solution

Answer

The velocity is 0 at \( t = 1 \) and \( t = 3 \) seconds. The corresponding accelerations are \(-6\) m/s² and \(6\) m/s², respectively.

Step 1: Find the velocity function

The velocity function is the derivative of the position function \( s(t) = t^{3} - 6t^{2} + 9t \). Differentiate \( s(t) \) with respect to \( t \):\[ v(t) = \frac{d}{dt}(t^3 - 6t^2 + 9t) = 3t^2 - 12t + 9 \].

Step 2: Set the velocity to zero to find critical points

To determine when the velocity is 0, solve the equation \( v(t) = 3t^2 - 12t + 9 = 0 \). First, simplify the equation by dividing all terms by 3:\[ t^2 - 4t + 3 = 0 \].

Step 3: Solve the quadratic equation

Factor the quadratic equation \( t^2 - 4t + 3 = 0 \) to find the values of \( t \):\[ t^2 - 4t + 3 = (t-3)(t-1) = 0 \]. So, \( t = 3 \) or \( t = 1 \).

Step 4: Find the acceleration function

The acceleration function is the derivative of the velocity function \( v(t) = 3t^2 - 12t + 9 \). Differentiate \( v(t) \) with respect to \( t \):\[ a(t) = \frac{d}{dt}(3t^2 - 12t + 9) = 6t - 12 \].

Step 5: Calculate the acceleration at critical points

Substitute \( t = 1 \) and \( t = 3 \) into the acceleration function to find the acceleration when the velocity is 0:For \( t = 1 \): \[ a(1) = 6(1) - 12 = -6 \] m/s².For \( t = 3 \):\[ a(3) = 6(3) - 12 = 6 \] m/s².

Key Concepts

Velocity Function

The concept of a velocity function is important in calculus when analyzing movement over time. In this exercise, the car's position at any time \( t \) is given by the position function \( s(t) = t^{3} - 6t^{2} + 9t \). To find how the car's speed changes, we need to calculate the velocity function. This requires us to take the derivative of the position function with respect to time \( t \).

• Taking the derivative involves using differentiation rules to find how \( s(t) \) changes over time.
• The velocity function \( v(t) \) is found by differentiating \( s(t) \): \( v(t) = \frac{d}{dt}(t^3 - 6t^2 + 9t) = 3t^2 - 12t + 9 \).
• This new function, \( v(t) = 3t^2 - 12t + 9 \), tells us how fast and in what direction the car is moving at any given \( t \).

Understanding velocity functions helps in determining when the car will stop or change direction, which is exactly when the velocity equals zero.

Acceleration Function

Acceleration refers to the rate at which velocity changes over time. It indicates how quickly a car speeds up or slows down. Here, we start from the already calculated velocity function, \( v(t) = 3t^2 - 12t + 9 \), and find its derivative to determine acceleration.

• The acceleration function is the derivative of the velocity function with respect to time \( t \).
• Thus, we differentiate \( v(t) \): \( a(t) = \frac{d}{dt}(3t^2 - 12t + 9) = 6t - 12 \).
• This acceleration function, \( a(t) = 6t - 12 \), gives us insights into how fast the car's velocity is changing at an instant \( t \).

When the velocity is zero, understanding the acceleration can tell us whether the car is speeding up or slowing down at that moment.

Derivative Calculation

The process of finding derivatives is essential in calculus for understanding how one quantity changes with another. In this context, it helps us determine both the velocity and acceleration functions.

• To compute a derivative, apply basic rules of differentiation such as the power rule.
• For a simple power function \( f(t) = t^n \), the derivative \( \frac{d}{dt}[t^n] = nt^{n-1} \).
• Using this rule, we calculate the derivative of the position function \( s(t) = t^{3} - 6t^{2} + 9t \), term by term.
• This gives us the velocity function, which we then differentiate again to find acceleration.

Mastering derivative calculation is crucial because these functions help predict and understand real-world phenomena such as motion, making them indispensable tools in physics and engineering.