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) Use standard gravity, a = 9.80665 m/s 2, for equations involving the Earth's gravitational force as the acceleration rate of an object. Find an algebraic formula, s(t), for the position of the person at time t, assuming that s(0) = 0. Explain your thinking.

Due to the current economic climate, unfortunate circumstances and the harsh reality of a super competitive industry, we have operated at a continued loss for a sustained period, this combined with not taking a wage has led myself and Elisha to re-evaluate the business while also doing what we must for our family and children.In this section, students will apply what they have learned about distance and displacement to the concepts of speed and velocity. For what values of t is the velocity of the ball positive? What does this tell you about the motion of the ball on this interval of time values? 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. On the left-hand axes provided in Figure 4.1, sketch a labeled graph of the velocity function v(t) = 3. Note that while the scale on the two sets of axes is the same, the units on the right-hand axes differ from those on the left. The right-hand axes 209 will be used in question (d).

The position vector from the origin of the coordinate system to point P is \(\vec{r}(t)\). In unit vector notation, introduced in Coordinate Systems and Components of a Vector, \(\vec{r}\)(t) is There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we will look at time, speed, and velocity to expand our understanding of motion. BL] [OL] Before students read the section, ask them to give examples of ways they have heard the word speed used. Then ask them if they have heard the word velocity used. Explain that these words are often used interchangeably in everyday life, but their scientific definitions are different. Tell students that they will learn about these differences as they read the section. 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 inHere we are making the implicit assumption that s(0) = 0; we will further discuss the different possibilities for values of s(0) in subsequent study. If only the average velocity is of concern, we have the vector equivalent of the one-dimensional average velocity for two and three dimensions: What does it mean to antidifferentiate a function and why is this process relevant to finding distance traveled? The runners are 6km "Left" and 5km "Right", therefore, the distance between the two is 11km, I'm not much of a fan of Km/h, so for calculations sake let's convert Km/h to M/s, this is only for me, doing the calculations in Hours and minutes is all fine, but I'm in love with metric.