Problem 38 - Calculus Early Transcendentals

Question

\(37-38\) A particle moves along a straight line with equation of motion \(s=f(t),\) where \(s\) is measured in meters and \(t\) in seconds. Find the velocity and the speed when t \(=5\) $$f(t)=t^{-1}-t$$

Step-by-Step Solution

Answer

Velocity at \( t = 5 \) is \(-\frac{26}{25}\) m/s; speed is \(\frac{26}{25}\) m/s.

Step 1: Calculate the Velocity Function

The velocity function is the derivative of the position function with respect to time. Given the position function is \( f(t) = t^{-1} - t \), find \( v(t) = f'(t) \). Compute the derivative: \( f'(t) = \frac{d}{dt}(t^{-1}) - \frac{d}{dt}(t) = -t^{-2} - 1 \). Thus, the velocity function is \( v(t) = -t^{-2} - 1 \).

Step 2: Evaluate the Velocity at t=5

Substitute \( t = 5 \) into the velocity function to find the velocity when \( t = 5 \). Use \( v(5) = -5^{-2} - 1 \). Calculate \( v(5) = -\frac{1}{25} - 1 = -\frac{1}{25} - \frac{25}{25} = -\frac{26}{25} \). Therefore, the velocity at \( t = 5 \) is \(-\frac{26}{25} \text{ m/s} \).

Step 3: Determine the Speed at t=5

Speed is the absolute value of the velocity. We found the velocity at \( t = 5 \) to be \(-\frac{26}{25}\). Therefore, the speed is \(|-\frac{26}{25}| = \frac{26}{25} \text{ m/s} \).

Key Concepts

The Derivative

In calculus, the derivative plays an essential role in understanding how quantities change. When we talk about the derivative of a function, we are looking at the rate at which the function's value changes concerning its variable.
For instance, if a particle is moving along a line, the derivative of its position function with respect to time gives us the velocity function, which describes how fast the position of the particle changes over time.

• The position function, often denoted as \( s(t) \) or \( f(t) \), tells us the location of a particle at any time \( t \).
• The derivative of the position function, \( f'(t) \), is the velocity function \( v(t) \).

In our exercise, we have \( f(t) = t^{-1} - t \). To find the velocity function, \( v(t) \), we calculated the derivative: \( f'(t) = -t^{-2} - 1 \). This calculation shows how the position changes over time, which directly relates to the particle’s velocity.

Velocity

Velocity is a vector quantity that refers to the rate of change of position with respect to time, alongside the direction of motion. Unlike speed, velocity can be negative, indicating movement in the opposite direction.

• The velocity function \( v(t) \) is derived from the position function by taking its derivative.
• A positive velocity indicates motion in the positive direction, while a negative velocity indicates the opposite direction.

For the given function \( f(t) = t^{-1} - t \), its derivative, \( v(t) = -t^{-2} - 1 \), gives the velocity. When evaluated at \( t = 5 \), the velocity is \( v(5) = -\frac{26}{25} \text{ m/s} \). This negative sign tells us that the particle is moving backward.

Speed

Speed is closely related to velocity but lacks direction. It is the absolute value of velocity, making it a scalar quantity that only considers magnitude. This provides the rate of movement without considering direction.

• To find the speed from velocity, we take the absolute value: \( \text{speed} = |v(t)| \).
• Speed is always non-negative, as it does not indicate the direction of motion.

With the velocity at \( t = 5 \) calculated as \( -\frac{26}{25} \text{ m/s} \), the speed becomes \( \frac{26}{25} \text{ m/s} \). This lets us know how fast the particle is moving, regardless of direction.