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Page 37 Position, Velocity and Acceleration Recall that when we differentiate the position function of an object, we obtain its _________ function. Now, if we differentiate the velocity function, we obtain the objects ______________ function. The following diagram is the relationship of the three functions: position, velocity and acceleration: Position Velocity Acceleration Example: Assume a car is traveling along a straight line with the position function given as s(t) = t 3 – 6t 2 + 9t. 1. Find the position of the car at t = 0s, 1s, 2s, 3s, 4s and 5s. 2. Find the velocity function of the car. 3.Find the velocity of the car at t = 0s, 1s, 2s, 3s, 4s and 5s. 4. Find the acceleration function of the car. 5. What is the acceleration of the car at t = 0s, 1s, 2s, 3s, 4s and 5s. As you recall, if the car has a positive velocity at a certain time, it means the car is moving _______ or _______ or _______; if the car has a negative velocity, it means the car is moving _______ or _______ or _______. So, what do positive and negative acceleration tell us about the movement of the car? Answer: ts(t)s(t)v(t)v(t)a(t)a(t) 0 0.53.1253.75–9 1 1.53.375–2.25–3 2 2.50.625–2.253 3 3.50.8753.759 4 4.510.12515.7515 5 If the car is moving to the right (i.e., velocity is positive), then a positive acceleration means the car is ___________ and a negative acceleration means the car is ____________. If the car is moving to the left (i.e., velocity is negative), then a positive acceleration means the car is ___________ and a negative acceleration means the car is ___________. The car is moving to the right (since the position is getting larger and velocity is positive), the negative acceleration means the car is __________ because |velocity| is getting smaller. The car is moving to the left (since the position is getting smaller and velocity is negative), the negative acceleration means the car is __________ because |velocity| is getting larger. The car is moving to the left (since the position is getting smaller and velocity is negative), the positive acceleration means the car is __________ because |velocity| is getting smaller. The car is moving to the right (since the position is getting larger and velocity is positive), the positive acceleration means the car is __________ because |velocity| is getting larger.

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Page 38 Throwing an Object Directly Upward and Dropping an Object When we throw an object, say a ball, directly upward, we must exert a force with some velocity, to make it go up. Of course, what makes it come back down is the _____________ force (or ____________ pull). The function used to described the position of the ball in air (from the moment the ball is let go to the moment it lands on the ground) is the height of ball relative to the ground and it can be written as: h(t) = –½g t 2 + v 0 t + h 0 where g is the gravitational acceleration constant on earth which is 9.8 m/s 2 (or 32 ft/s 2 ), and v 0 is the initial velocity and h 0 is the initial height. Since g is a constant, either 9.8 or 32, so the height of the ball can be written as h(t) = –4.9t 2 + v 0 t + h 0 if its measured in meters or h(t) = –16t 2 + v 0 t + h 0 if its measure in feet. Example 1 You throw a ball upward with velocity 3.5 m/s and assume your hand (the one that is throwing the ball) is 1.4 m above ground when the ball is thrown, a) At what time does the ball reach the maximum height and what is the maximum height? b) At what time does the ball hit the ground and what is the velocity when it hits the ground? Example 2 Standing on the edge of a canyon, you took out a coin from your pocket, and the fun of it, you simply let go of it and let it drop to the bottom of the canyon. Assume the distance between the hand and the bottom of the canyon is 1,600 ft and assume there is no air resistance, a) How long does it take the coin to hit the ground? What is the velocity when the coin hits the ground? b) If you drop the coin with a velocity of 72ft/s, how long does it take the coin to hit the ground? What is the velocity when the coin hits the ground?

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Page 39 Derivative at a Point = 0 Besides at the Local Maximum/Minimum Points? Recall that we have mentioned that the derivative of a function at a local maximum/minimum points is __. This statement is not totally accurate. There are functions where the derivative at the local maximum/minimum points fails to exist (DNE) and there are functions with points with 0 derivative without being a local maximum/minimum point. How can that be? Lets see the graph on the right: –7–6–5–4–3–2 –10 1 2 3 4 567 3 2 1 –1 –2 –3 1. At what points is the derivative = 0? 2.At what points is the derivative DNE? Lets call a point where the derivative fail to exist a non-differentiable point (NDP). So both (0, 2) and (4, –3) are NDPs. At (0, 2) we cant draw a tangent line, however, at (4, –3), we actually can draw the tangent line. How can that be? Its because we have to look at the points on either side of and near the NDP, and ask ourselves what kind of tangent lines can we draw at these points before they get to the NDP. As we can see that are points, even if they are not local max/min points, its derivative is 0, e.g., (_, _). On the other hand, there are points, even if they are local max/min points, its derivative is DNE, e.g., (_, _) and (_, _). We can see that, as a point nears (0, 2), if it is on the left side of (0, 2), the slope of the tangent line drawn on it is a positive number that almost wont change even it gets to (0,2). However, if a point is on the right side of (0, 2), the slope of the tangent line drawn on it is a _________ number that almost wont change even it gets to (0,2). So we cant draw the tangent line and hence, the derivative at this point is _____. On the other hand, as a point nears (4, –3), if it is on the left side of (4, –3), the slope of the tangent line drawn on it has a negative slope but the line is very steep and getting steeper as it moves closer to (4, –3). However, if it is on the right side of (4, – 3), the slope of the tangent line drawn on it has a _________ slope but the line is very steep and getting steeper as it moves closer to (4, –3). Therefore, when these points finally get to (4, –3), the tangent line can be drawn is a ____________ tangent line. Of course, the derivative at this point is DNE, not because we cant draw a tangent line, but because the slope of a vertical (tangent) line is ______________. Pictures with points with a horizontal tangent line: Pictures with points with a vertical tangent line:

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