Velocity Progear ROGUE PB 9.0 SERVICE BAG, Black

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Find an algebraic formula, s(t), for the position of the person at time t, assuming that s(0) = 0. Explain your thinking. Changes were made to the original material, including updates to art, structure, and other content updates. Using the graph of y = v(t) provided in Figure 4.6, find the exact area of the region under the velocity curve between t = 1 2 and t = 1. What is the meaning of the value you find? begin{align*} \vec{r}(t_{1}) &= 6770 \ldotp \; km\; \hat{j} \\[4pt] \vec{r}(t_{2}) &= 6770 \ldotp \; km (\cos (-45°))\; \hat{i} + 6770 \ldotp \; km (\sin(−45°))\; \hat{j} \ldotp \end{align*}\] Use standard gravity, a = 9.80665 m/s 2, for equations involving the Earth's gravitational force as the acceleration rate of an object.

It is always a straight line path and can never be a curve, zig- zag or some irregular path joining the initial and final positions of the body that is why it is also defined as the shortest path length travelled by the body joining the initial and final positions of the body. For what values of t is the position function s increasing? Explain why this is the case using relevant information about the velocity function v.In the previous chapter we found the instantaneous velocity by calculating the derivative of the position function with respect to time. We can do the same operation in two and three dimensions, but we use vectors. The instantaneous velocity vector is now A ball is tossed vertically in such a way that its velocity function is given by v(t) = 32 − 32t, where t is measured in seconds and v in feet per second. Assume that this function is valid for 0 ≤ t ≤ 2. On the right-hand axes provided in Figure 4.1, sketch a labeled graph of the position function y = s(t).

Now, the analogy is, that if A and B where too meet, they would do so at the same time, A cannot collide with B and then B collides with A, its a mutual collision between the two runners, both must be travveling for ~39 minutes before they meet eachover in the run, it seems rather counter intuitive at first since the ~39 minutes is the time taken for A, moving at the relative velocity of A+B too meet B, who in this case, for maths, is stationary, but it does work out, since when you use the "Normal" velocities of both, rather then having one going "Super fast" and the other being "Stationary" you find that it does in fact take them 39 minutes to both collectively cover the 11km (A does, say 6 of those Kilometers, and B does 5) Bart walks west at 5 m/s for 3 seconds, then turns around and walks east at 7 m/s for 1 second. We can treat the eastward movement as "negative movement west," so total displacement = (5 m/s west)(3 s) + (-7 m/s west)(1 s) = 8 meters. Total time = 4s. Average velocity = 8 m west / 4s = 2 m/s west. Suppose that an object moving along a straight line path has its velocity v (in meters per second) at time t (in seconds) given by the piecewise linear function whose graph is pictured in Figure 4.8. We view movement to the right as being in the positive direction (with positive velocity), while movement to the left is in the negative direction. Suppose In particular, when velocity is positive on an interval, we can find the total distance traveled by finding the area under the velocity curve and above the t-axis on the given time interval. We may only be able to estimate this area, depending on the shape of the velocity curve. Answer the same questions as in (c) and (d) but instead using the interval [0, 1]. (f) What is the value of s(2) − s(0)? What does this result tell you about the flight of the ball? How is this value connected to the provided graph of y = v(t)? Explain. C

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newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\) Area under the graph of the velocity function In Preview Activity 4.1, we encountered a fundamental fact: when a moving object’s velocity is constant (and positive), the area under the velocity curve over a given interval tells us the distance the object traveled. As seen at left in Figure 4.2, if we consider an object

vec{v} (t) = \lim_{\Delta t \rightarrow 0} \frac{\vec{r} (t + \Delta t) - \vec{r} (t)}{\Delta t} = \frac{d \vec{r}}{dt} \ldotp \label{4.4}\] When we look at the three-dimensional equations for position and velocity written in unit vector notation, Equation \ref{4.2} and Equation \ref{4.5}, we see the components of these equations are separate and unique functions of time that do not depend on one another. Motion along the x direction has no part of its motion along the y and z directions, and similarly for the other two coordinate axes. Thus, the motion of an object in two or three dimensions can be divided into separate, independent motions along the perpendicular axes of the coordinate system in which the motion takes place.When the velocity of a moving object is positive, the object’s position is always increasing. While we will soon consider situations where velocity is negative and think about the ramifications of this condition on distance traveled, for now we continue to assume that we are working with a positive velocity function. In that setting, we have established that whenever v is actually constant, the exact distance traveled on an interval is the area under the velocity curve; furthermore, we have observed that when v is not constant, we can estimate the total distance traveled by finding the areas of rectangles that help to approximate the area under the velocity curve on the given interval. Hence, we see the importance of the problem of finding the area between a curve and the horizontal axis: besides being an interesting geometric question, in the setting of the curve being the (positive) velocity of a moving object, the area under the curve over an interval tells us the exact distance traveled on the interval. We can estimate this area any time we have a graph of the velocity function or a table of data that tells us some relevant values of the function. In Activity 4.1, we also encountered an alternate approach to finding the distance traveled. In particular, if we know a formula for the instantaneous velocity, y = v(t), of the moving body at time t, then we realize that v must be the derivative of some corresponding position function s. If we can find a formula for s from the formula for v, it follows that we know the position of the object at time t. In addition, under the assumption that velocity is positive, the change in position over a given interval then tells us the distance traveled on that interval. For a simple example, consider the situation from Preview Activity 4.1, where a person is walking along a straight line and has velocity function v(t) = 3 mph. As pictured in It is with a heavy heart and much consideration we have decided that loadout, as we all know it, will cease trading with immediate effect.