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Chapter 6: Problem 31

A rock is dropped from the top of a 400 -foot cliff. Its velocity at time \(t\) seconds is \(v(t)=-32 t\) feet per second. Find the displacement of the rock during the time interval \(2 \leq t \leq 4\).

Short Answer

Expert verified
The displacement is -192 feet.

Step by step solution

01

Understand the Given Information

The initial height of the cliff is 400 feet, and the velocity of the rock as a function of time is provided as \(v(t) = -32t\) feet per second. We're asked to find the displacement of the rock from time \(t = 2\) to \(t = 4\).
02

Set Up the Displacement Integral

Displacement can be found by integrating the velocity function over the given time interval. We need to evaluate the integral: \[\text{Displacement} = \ \ \int_{2}^{4} v(t) \, dt = \ \ \int_{2}^{4} -32t \, dt\]
03

Find the Indefinite Integral

Calculate the indefinite integral of the velocity function: \(\int -32t \, dt = -16t^2 + C\)
04

Evaluate the Definite Integral

To find the displacement, evaluate the definite integral from \(t = 2\) to \(t = 4\): \[\text{Displacement} = \ \ [-16t^2]_{2}^{4} \] \[ = -16(4)^2 - (-16(2)^2) \ \ = -16(16) - (-16(4)) \ \ = -256 + 64 \ \ = -192 \text{ feet}\]
05

Interpret the Result

The displacement of the rock over the time interval \(2 \leq t \leq 4\) is -192 feet. The negative sign indicates the rock has moved downward.

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Key Concepts

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

velocity function
In calculus, the **velocity function** describes how the velocity of an object changes over time. For instance, in our rock displacement problem, the velocity function is given as \( v(t) = -32t \). This tells us the rock's velocity at any time \( t \) seconds after being dropped. The negative sign indicates that the rock is moving downwards, as velocity is a vector quantity and has direction. The rate of change of the rock's position, or its speed, increases linearly with time because it is affected by gravity.
definite integral
A **definite integral** is used to calculate the area under the curve of a function within a specified interval. For our problem, we use the definite integral of the velocity function to find the displacement of the rock between times \( t = 2 \) and \( t = 4 \). The definite integral is set up as:
\[ \text{Displacement} = \int_{2}^{4} -32t \, dt \]
By evaluating this integral, we can determine the total distance the rock has moved (or its displacement) within this time period. Note that the integral gives us a net value, which accounts for direction, resulting in a negative value as the rock moves downward.
indefinite integral
An **indefinite integral** represents the antiderivative of a function and includes a constant of integration \( C \). For the velocity function \( v(t) = -32t \), the indefinite integral is found as follows:
\[ \int -32t \, dt = -16t^2 + C \]
Here, the \( -16t^2 \) term is the antiderivative of \( -32t \), and \( C \) represents an arbitrary constant. Although \( C \) is part of the indefinite integral, it cancels out when computing the definite integral over a specific interval, making it unnecessary in the evaluation of displacement.
displacement
In physics and calculus, **displacement** refers to the change in position of an object. It's a vector quantity, meaning it has both magnitude and direction. For the rock displacement problem, we computed the displacement over the interval \( 2 \leq t \leq 4 \) by evaluating the definite integral of the velocity function:
\[ \text{Displacement} = [ -16t^2 ]_{2}^{4} = -16(4)^2 - (-16(2)^2) = -256 + 64 = -192 \text{ feet} \]
This result means the rock moved 192 feet downward from time \( t = 2 \) to \( t = 4 \). The negative sign indicates the downward movement, which aligns with our velocity function being negative.

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