13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.

b. Find the displacement over the given interval.

v(t) = 50e^−2t on [0, 4]

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Identify the displacement as the definite integral of the velocity function over the given time interval. Displacement is given by \(\int_{a}^{b} v(t) \, dt\), where \(a=0\) and \(b=4\) in this problem.

Write down the integral to find displacement: \(\int_{0}^{4} 50 e^{-2t} \, dt\).

Recall the integral formula for an exponential function: \(\int e^{kt} \, dt = \frac{1}{k} e^{kt} + C\). Here, \(k = -2\).

Apply the integral formula to \(50 e^{-2t}\): the antiderivative is \(50 \times \frac{1}{-2} e^{-2t} = -25 e^{-2t}\).

Evaluate the definite integral by substituting the limits: calculate \([-25 e^{-2t}]\) from \(t=0\) to \(t=4\), which means computing \(-25 e^{-8} - (-25 e^{0})\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Displacement and Velocity Relationship

Displacement represents the change in position of an object and is found by integrating the velocity function over a given time interval. Since velocity is the rate of change of position, integrating velocity with respect to time gives the net displacement.

Definite Integral

A definite integral calculates the accumulated quantity, such as displacement, over a specific interval. For velocity functions, the definite integral from time a to b gives the total displacement between those times.

Exponential Decay Function

The velocity function v(t) = 50e^(-2t) is an exponential decay, meaning velocity decreases rapidly over time. Understanding how to integrate exponential functions is essential to find displacement accurately.

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.b. Find the displacement over the given interval. v(t) = 50e^−2t on [0, 4]