Question

A bird is flying due east. Its distance from a tall building is given by \(x(t) =\) 28.0 m \(+\) (12.4 m/s)\(t\) - (0.0450 m/s\(^3)t^3\). What is the instantaneous velocity of the bird when \(t =\) 8.00 s?

Short answer

The instantaneous velocity at \(t = 8.00\) s is \(3.76\ m/s\).

Step-by-step solution

Understand the Distance Function

The given function is \(x(t) = 28.0 + 12.4t - 0.0450t^3\). This represents the distance \(x\) as a function of time \(t\). The units are in meters and seconds.

Differentiate to Find Velocity

The instantaneous velocity is the derivative of the distance function with respect to time. So, find \(v(t) = \frac{dx(t)}{dt}\). Using differentiation rules:\[v(t) = \frac{d}{dt}(28.0) + \frac{d}{dt}(12.4t) - \frac{d}{dt}(0.0450t^3)\]This simplifies to:\[v(t) = 0 + 12.4 - 0.135t^2\]Therefore, \(v(t) = 12.4 - 0.135t^2\).

Evaluate Velocity at t = 8.00 s

Plug \(t = 8.00 s\) into the velocity function:\[v(8) = 12.4 - 0.135(8)^2\]First, calculate \(8^2 = 64\). Then:\[v(8) = 12.4 - 0.135(64)\]\(0.135 \times 64 = 8.64\), so:\[v(8) = 12.4 - 8.64 = 3.76\]

Conclusion: Report the Instantaneous Velocity

The instantaneous velocity of the bird at \(t = 8.00\) seconds is \(3.76\ m/s\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that lets us find the rate of change of a function. In simple terms, when we're given a function that tells us how a quantity changes over time, differentiation helps us determine how fast that quantity is changing *right at any specific instant*.
For the problem with the bird, the distance function given is in terms of time, and we are tasked with finding the instantaneous velocity. To do this, we perform differentiation on the distance function.

• The derivative of a constant term, like 28.0, is zero because constant values don’t change hence have no rate of change.
• The derivative of a linear term, such as 12.4t, is simply the coefficient in front of t, which is 12.4.
• For the cubic term, 0.0450t³, we apply the power rule. Multiply the exponent (3) by the coefficient (0.0450) to get 0.135, and decrease the exponent by one to t², resulting in 0.135t².

The result is a new function that gives the velocity of the bird at any given time. This derivative is precisely the velocity function, important for predicting how quickly the bird is moving at any point in time.

Distance Function

A distance function is used to describe how far an object has moved over time. In physics, it's crucial to visualize how objects travel in space, often relative to another point.
In our example, the bird's distance from a tall building is given by the function \(x(t) = 28.0 + 12.4t - 0.0450t^3\). This formula tells us:

• At \(t = 0\), the bird is already 28.0 meters away from the building. This is a starting fixed point known as an initial condition.
• The term 12.4t indicates that as time progresses, there's a straightforward increase in distance attributed to the bird flying east at a steady initial speed of 12.4 m/s.
• The \(-0.0450t^3\) term suggests a correction factor for changes in speed, accounting for possible deceleration or external forces impacting the bird's flight.

Understanding distance functions helps us interpret how objects like birds move, using mathematical models to predict their paths and velocities. This approach is essential for solving physics problems about motion.

Physics Problem Solving

Physics problems often involve translating real-world situations into mathematical expressions. Here, we have a bird's motion described mathematically, asking us to find a specific instant of velocity.
Solving such problems requires a structured approach:

• Understand the problem: Identify what you're asked for and what information you have, such as the bird's initial position and motion description.
• Use relevant formulas: Utilize calculus, specifically differentiation, to shift from distance to velocity functions. Recognize which terms reflect constant motion, acceleration, or deceleration.
• Calculate precisely: With the obtained velocity function, plug in specific values (e.g., \(t = 8.00 s\)) to solve for the desired quantity, ensuring accurate arithmetic.

Such problem-solving practices build a fundamental backbone for interpreting mathematical relationships in physics. Ultimately, they assist in visualizing dynamic systems, allowing for practical predictions and decisions based on mathematical clarity.