Piecewise velocity The velocity of a (fast) automobile on a straight highway is given by the function v(t)={3t if 0≤t\u003c2060 if 20≤t\u003c45240−4t if t≥45 v(t)= \begin{cases}3 t & \text { if } 0 \leq t\u003c20 \\ 60 & \text { if } 20 \leq t\u003c45 \\ 240-4 t & \text { if } t \geq 45\end{cases} v(t)=⎩⎨⎧3t60240−4t if 0≤t\u003c20 if 20≤t\u003c45 if t≥45 where is measured in seconds and v has units of m/s. d. What is the position of the automobile when t=75?

Recall that the position function $$s(t) is $$the integral of the velocity function $$v(t)$$ with respect to time, plus an initial position constant: $$s(t) = s(0) + \$$int_0$$^t v(\tau) \, d\tau. $$Since the velocity $$v(t) is $$given as a piecewise function, break the integral from 0 to 75 into three parts corresponding to the intervals of $$v(t)$$: from 0 to 20, from 20 to 45, and from 45 to 75. Set up the integral for each interval: - For $$0 \leq t \u003c 20$$, integrate $$3t$$ with respect to $$t$$ from 0 to 20. - For $$20 \leq t \u003c 45$$, integrate the constant velocity 60 from 20 to 45. - For $$t \geq 45$$, integrate $$240 - 4t$$ from 45 to 75. Calculate each definite integral separately to find the displacement over each time interval. Then, sum these displacements to find the total change in position from $$t=0 to$$ $$t=75. If $$the initial position $$s(0) is $$known (often assumed to be zero if not given), add it to the total displacement to find the position $$s(75) at $$time $$t=75.$$

Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Ch. 6 - Applications of Integration

Problem 6.1.27d

Chapter 6, Problem 6.1.27d

Piecewise velocity The velocity of a (fast) automobile on a straight highway is given by the function
$v(t)= $\begin{cases}$3 t & $\text$ { if } 0 $\leq$ t<20 \\ 60 & $\text$ { if } 20 $\leq$ t<45 \\ 240-4 t & $\text$ { if } t $\geq$ 45$\end{cases}$$
where is measured in seconds and v has units of m/s. d. What is the position of the automobile when t=75?

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Recall that the position function $s(t)$ is the integral of the velocity function $v(t)$ with respect to time, plus an initial position constant: $s(t) = s(0) + \int_0^t v(\tau) \, d\tau$.

Since the velocity $v(t)$ is given as a piecewise function, break the integral from 0 to 75 into three parts corresponding to the intervals of $v(t)$: from 0 to 20, from 20 to 45, and from 45 to 75.

Set up the integral for each interval: - For $0 \leq t < 20$, integrate \$3t$ with respect to $t$ from 0 to 20. - For $20 \leq t < 45$, integrate the constant velocity 60 from 20 to 45. - For $t \geq 45$, integrate $240 - 4t$ from 45 to 75.

Calculate each definite integral separately to find the displacement over each time interval. Then, sum these displacements to find the total change in position from $t=0$ to $t=75$.

If the initial position $s(0)$ is known (often assumed to be zero if not given), add it to the total displacement to find the position $s(75)$ at time $t=75$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Piecewise Functions

A piecewise function is defined by different expressions over distinct intervals of the independent variable. Understanding how to evaluate and interpret these functions on each interval is essential, especially when the function changes behavior at specific points, as with the velocity function given.

Relationship Between Velocity and Position

Velocity is the derivative of position with respect to time, so position can be found by integrating velocity over time. To find the position at a certain time, you integrate the velocity function from the initial time to that time, accounting for changes in velocity across intervals.

Definite Integration of Piecewise Functions

When integrating a piecewise function, you must split the integral at the points where the function definition changes. This means calculating the integral over each interval separately and summing the results to find the total change in position.

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