Position,+Velocity,+Acceleration

Definitions: Position- the exact location of a given object at a given time. Velocity- change in the objects position over a certain amount of time; speed of a given object Acceleration- defined by the change of an objects velocity over a certain period of time

Position: because position is defined as the exact location of a given object, a change in its position over a period of time would result in a velocity; therefore one way of finding a position is to take the integral of a velocity.

To find exact location of an object, you can algebraically solve the kinematic equation d=v(initial)*t + (1/2)at^2 for position:

p= p(initial) + v(initial)*t + (1/2)at^2

p- position v-velocity t- time at which your looking to find the object's postiona- objects acceleration at time, t

A visual image of position is as shown below:

 As time (in seconds) pass, the postion (in meters) of the car changes.

Position is sometimes desribed as the area under a V(t) graph, for example:

If f(x) reprersents velocity, then in this particular diagram, then the value of 'S' will represent the displacemnet between point 'a' and point 'b'. If you were looking for a single postion you could take the area underneath the curve between the x-value 0 and that value.

Velocity: As defined above is a vector quantity refered to as the change in a position of an object in a given amount if time. Therefore if velocity is defined as a change in position, it can be found by taking the derivative of position and/or the antiderivative of acceleration.

Below is the equations of velocity as a function of acceleration and position: Looking at the diagram of the car, it shows that the car is going at a constant velocity of 10 m/s; which is simpler concept to understand until the car's velocity begins to change. Graphically: You can the slope (velocity) goes up at a constant rate and is straight.

When looking at the following diagram, we can see an increase in the velocity of the the car, also known as acceleration.

Graphically: In this graph you can see that the distance between each position is getting increasingly larger as time increases. The velocity is increasing, causing the curve to the graph. Acceleration: Is defined as the change in velocity. It can be found by taking the derivaive of velocity, or the second derivative of position.

a(t)=p'(t)dx

If we look back at the original diagram of the car we can see that it velocity is constant, as we found before.



Since the velocity as we said was constant, we know the acceleration is zero because their is no change in the cars velocity.

When looking at this diagram we can see that the car is changing its velocity because the rate at which the distance is changing is increasing each time interval, therefore if the velocity of the car is changing then we know that the car is accelerating due to the definition of acceleration.

To better help connect the three ideas look at the diagram below:



Practice Problems: *Answers to question 1-3 are below in red *

#1 The position function of a particle moving on a straight line is s(t)=2t^3 - 10t^2 +5. Find: (a). Position (b). Instantaneous Velocity (c). Acceleration (d). Speed of the particle at t=1.

#2 The velocity function of a moving particle is v(t)= (t^3 /3) - 4t^2 =16t -64 for zero is less than or equal to t and t is less than or equal to 7.

What is the minimum and maximum acceleration of the particle on the given interval?

#3 The graph of the velocity function is shown in Figure 9.3-2.



1. When is the acceleration 0?

2. When is the particle moving to the right?

3. When is the speed the greatest? Answers: #1 a. s(1)= 2(1)^3 -10(1)^2 +5= -3 b. v(t)= s'(t)= 6t^2 -20t v(1)= 6(1)^2 -20(1)= -14 c. a(t)= v'(t)= 12t -20 a(1)= 12(1) -20= -8 d. Speed= abs(v(t))= abs(v(1))= 14 1. The minimum acceleration occurs at t=4 and a(4)=0 2. The maximum acceleratin occurs at t=0 and a(0)=16 #3 1. a(t)= v'(t) and v'(t) is the slope of the tangent to the graph of v. At t=1 and t=3, the slope of the tangent is 0. 2. For 2<t<4, v(t) >0. Thus the particle is moving to the right during 2<t<4. 3. Speed = abs(v(t)) at t=1, v(t)= -4. Thus, speed at t=1 is abs(-4)= 4 which is the greatest speed for 0 is less than or equal to t and t is less than or equal to 4. <span style="color: #ff0000; font-family: 'Comic Sans MS',cursive;">
 * 1) 2. See Figure 9.3-1. The graph of a(t) indicates that:

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