EGR 1980 Recitation Worksheet 5 Constant Acceleration Kinematic Equations for One Dimensional Motion H Griffith 10 March 2014 Introduction In last week s recitation we defined the basic kinematic quantities and discussed how velocity varied in time for the case of one dimensional motion under constant acceleration Recall that for constant acceleration velocity varied linearly over time increasing or decreasing depending upon the direction of the acceleration vector relative to the velocity vector If an object is decreasing its speed over time such as the case of a braking car the direction of the acceleration vector is opposite that of the velocity vector and the velocity decreases linearly over time from its initial value If an object is increasing its speed over time the two vectors align and the velocity increases linearly over time Today we ll examine how the position of the object varies under this identical type of motion Additionally we ll discuss some basic geometrical relationships between the graphs of each of the three kinematic quantities Variation in Position Over Time Under Constant Acceleration As we discussed last week the velocity of an object experiencing constant acceleration varies in time according to the equation below v t v o at As we are currently discussing in lecture the above expression is written using function notation Namely the left side of the equation v t indicates that the velocity represented by the variable v varies over time represented by the variable t We may demonstrate that this relationship is in fact a function by subjecting it to the vertical line test Namely note that any vertical line drawn through a velocity versus time graph intersects the velocity curve at only one point This is shown below for the specific case of a car braking at constant acceleration v m s t s You should also recognize that for the graph above the so called x y plane here represented as the t v plane is only shown in the first quadrant The physical meaning of this choice indicates that only positive values of time are valid in the model and that the value of the velocity at each of these points in time is also positive This restriction on only describing velocity at positive values of time serves to restrict the domain of the function Namely the domain of the above function may be written as follows D 0 t s Similarly we notice that for a car braking from some positive velocity to zero velocity the range of the function may be written as follows R 0 v o The above domain restrictions are especially important since the underlying mathematical model ie v t v o at will give values for points in time which are outside of the domain You try A car accelerates from a velocity of 30 m s to 50 m s over a 10 second interval Write an expression for the variation in the velocity as a function of time over this interval draw the corresponding graph and write the domain and range of your resulting function using interval notation In EGR 1010 you will use calculus in order to show that the displacement over an object under constant acceleration versus time is given by the following formula 1 x t x o vt a t 2 2 Note that the above expression is no longer linear but is instead quadratic As we will discuss in lecture graphs of quadratic functions form a class of U shaped curves known as parabolas An example of the graph of a parabola representing the displacement of an object increasing in speed ie a 0 is shown below 120 Displacement Meters 100 80 60 40 20 0 0 1 2 3 4 5 6 Time Seconds 7 8 9 10 Note that in the above graph the displacement at t 0 seconds is not 0 m The displacement at this point in time is referred to as the initial displacement and is represented in the above equation by xo The graph of an object decreasing in speed ie a 0 is also quadratic but does not have the same curvature as that shown above An example of such a graph is shown below 16 Displacement Meters 14 12 10 8 6 4 2 0 0 5 1 1 5 2 2 5 3 Time Seconds 3 5 4 4 5 5 You should note that although both graphs are increasing functions of time they show differences in their shape Mathematically we say the two curves differ in their concavity Namely we say that the first curve is concave up while the second curve is concave down You should note that the only difference between the two represented motions is the sign of the acceleration From this observation we may state the following For an object in linear motion under constant acceleration the graph of the displacement is quadratic If the acceleration is positive ie speeding up the quadratic graph is concave up If the acceleration is negative ie slowing down the quadratic graph is concave down We may use these observations along with what was covered last week in order to draw graphs which describe both the position and velocity of the motion as shown in the example below Example 1 A car is moving 50 m s and steps on the brake pedal in order to come to a stop We may assume that the acceleration over the period of braking is constant Five seconds after beginning to stop it is observed to be traveling at 30 m s This information may be tabulated as follows time s 0 5 velocity m s 50 30 1 Using the relationship shown below calculate the magnitude of the acceleration of the car during braking a v t 2 Using your knowledge of vectors draw vectors representing both the initial velocity and constant acceleration 3 As was discussed above the velocity of an object under linear motion may be represented by the following equation v t v o at Based upon the information calculated above specify v o and a for the motion described in this problem and write an expression for the velocity as a function of time 4 Define the domain and range of your function above 5 Graph the acceleration versus time over the period of braking What is the slope of your resulting curve 6 Graph the velocity versus time over the period of breaking 7 Using the equation developed in 3 determine the time required for the car to reach a complete stop 8 Using the information presented in this sheet what time of curve is formed by graphing the displacement of the object versus time What is the concavity of the curve 9 As was discussed above the displacement of an object under linear motion may be represented by the following equation 1 2 x t x o v o t a t 2 If we define the point at which the car begins braking as our reference ie at t 0 s x o 0 m we may simplify the formula as shown below 1 2 x t v o t …