SPH 3U Grade 11 U Physics • This lesson will introduce you to equations of motion called kinematic equation • You will be able to use these equations to solve kinematic related questions SPH 3U Grade 11 U Physics What are we going to cover today? By the end of this lesson, you will be able to: Look back at v-t graphs The five equations Sample problems Homework SPH 3U Grade 11 U Physics Recall : • Slope of a d-t graph is v • Slope of a v-t graph is a • Area under a • Area under a v-t graph is Dd a-t graph is Dv velocity (m/s [U]) SPH 3U Grade 11 U Physics Slope = a vf vi area = Dd ti tf time (s) 3s What would we need in order to graph the acceleration of a car? Motion Detector + computer with specialized software slow- motion camera Some kind of scale / reference for distance Timer We would need to create a 𝒅˗𝒕 graph, determine instantaneous (slope of tangent) and average velocity, plot the 𝑣 data and make a 𝒗˗𝒕 graph. Then we would need to find the slope to determine the 𝒂𝒂𝒗 . Or…. If the raw data is provided, we could just use MATH (algebra) + 𝒐𝒖𝒓 𝑩𝑹𝑨𝑰𝑵 Five Key Uniform Acceleration Equations As shown in the previous lessons, graphical analysis is an important tool physicists to use to solve problems. However, sometimes we have enough information to solve problems algebraically. Algebraic methods tend to be quicker and more convenient than graphical analysis. When solving uniform acceleration problems, choose which equations to use based on the given and required variables of the problem. EQUATION 1 To be able to solve uniform acceleration problems, we need to derive algebraic equations from a change in velocity using a standard 𝒗˗𝒕 graph . • The 1st Equation stems from the area under the 𝒗˗𝒕 graph EQUATION 1 Y=intercept B = base (run) = ∆t l = length (run) = ∆t h = height (rise) = ∆v h = height (rise) = vi Like Terms EQUATION 2 • The 2nd Equation stems from the slope of the 𝒗˗𝒕 graph and rearranging for 𝑣𝑓 . • What does slope of 𝒗˗𝒕 tell us? 𝒗𝒇 − 𝒗𝒊 𝒂𝒂𝒗 = • Rearrange for 𝑣𝑓 ∆t X ∆t X ∆t 𝒂𝒂𝒗 (∆t ) = 𝒗𝒇 − 𝒗𝒊 + 𝑣𝑖 𝒂𝒂𝒗 (∆t ) + 𝒗𝒊 = 𝒗𝒇 + 𝑣𝑖 EQUATION 3 In order to produce the 3rd Equation, we substitute Equation 2 into Equation 1 for 𝒗𝒇 as shown below. EQUATION 3 𝑳𝒊𝒌𝒆 𝑻𝒆𝒓𝒎𝒔 Get rid of the brackets Via multiplication = ∆𝒅 = 1 𝟐 (2 𝒗𝒊 + 𝒂𝒂𝒗 ∆t ) ∆t 1 𝒗𝒊 ∆t + 𝒂𝒂𝒗 ∆t 𝟐 𝟐 EQUATION 4 & 5 Equations 4 and 5 are similarly produced via substitution rearranging Equation 2 for 𝒗𝒊 rearranging Equation 2 for ∆t. The 5 Kinematic Equations SP #1-3 p.38-39 It is important to note that these five equations are used only when an object undergoes acceleration, a change in velocity. If an object is travelling at a constant velocity, we simply use the equation for constant velocity from section 1.2. REMEMBER: there is only 5 possible variables in any of these problems. Displacement, time, initial velocity, final velocity, & average acceleration Problems in this course can only contain 4 of the possible 5 variables. Each of the 5 equations is therefore missing a different variable – this ensures that no matter what variable is missing, we have a formula we can use to solve the problem. This also means we can figure out what formula we need to use by looking at our givens OR determining which of the 5 variables listed above we DON’T HAVE SPH 3U Grade 11 U Physics Ex 1: A runner jogging west at 5.0 m/s accelerates to 12 m/s while covering 150 m. What is the acceleration of the runner? Sol’n: Given: vi = 5.0 m/s vf = 12 m/s Dd = 150 m Required:a Analysis: Look for equation with all of these variables. 2010 SPH 3U Grade 11 U Physics v2f v2i 2aDd a 2 2 v v f i 2 Dd 2 2 12 5 2(150) .39667 The acceleration is .40 m/s2 [W] 1.5 Homework Practice # 1,2 p.39 Questions #1-6 p.39