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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

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

Step by step solution

01

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.
02

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, \(v(t) = 12.4 - 0.135t^2\).
03

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:\[v(8) = 12.4 - 8.64 = 3.76\]
04

Conclusion: Report the Instantaneous Velocity

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 m/s. (a) How tall is the cliff? (b) If we ignore air resistance, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (\(\textbf{Fig. E2.44}\)). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_y-t\), \(v_y-t\), and \(y-t\) graphs for the motion.

Starting from the front door of a ranch house, you walk 60.0 m due east to a windmill, turn around, and then slowly walk 40.0 m west to a bench, where you sit and watch the sunrise. It takes you 28.0 s to walk from the house to the windmill and then 36.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your (a) average velocity and (b) average speed?

Cars \(A\) and \(B\) travel in a straight line. The distance of \(A\) from the starting point is given as a function of time by \(x_A(t) = \alpha{t} + \beta{t}^2\), with \(\alpha =\) 2.60 m/s and \(\beta =\) 1.20 m/s\(^2\). The distance of \(B\) from the starting point is \(x_B(t) = \gamma{t}^2 - \delta{t}^3\), with \(\gamma =\) 2.80 m/s\(^2\) and \(\delta =\) 0.20 m/s\(^3\). (a) Which car is ahead just after the two cars leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

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