PHYSICS STUDY GUIDE CHAPTER 3: MOTION WITH CONSTANT ACCELERATION TOPICS: Motion with constant Acceleration WHAT YOU MUST KNOW Be able to sketch a situation Be able to draw a motion diagram Be able to draw graphs such as: o Motion Diagram o Position vs. clock reading o Velocity vs. clock reading o Acceleration vs. clock reading Be able to write mathematical models: o That describes the position of an object that moves with constant acceleration. o That describes the velocity of an object that moves with constant acceleration. o That describes the acceleration of an object that moves with constant acceleration. Make predictions to find velocities, positions and displacements of an object moving with constant acceleration. Understand that in the position vs. clock reading graph the displacement of the object can be found by subtracting the final and initial position on the position axis. Understand that in the velocity vs. clock reading graph the displacement of the object can be found as the area bounded between the graph and the horizontal axis (shape of the trapezoid). CHAPTER SUMMARY Motion with constant acceleration is the motion of an object that moves in a straight line and its velocity changes constantly each second. PHYSICAL QUANTITIES ACCELERATION The change in velocity of an object, over the change in time vxf - vxi aX = tf - ti Symbol: aX Type of PQ: Vector (direction matters) Units: m/s2 Notes: o o o Acceleration is the slope of the Velocity vs. Clock reading graph. Acceleration can be positive or negative. Acceleration is NOT an indicator of an object slowing down or speeding up. In motion with constant acceleration we studied 8 representations: 1. Description with words A battery operated cart accelerates at a constant rate of +1 m/s2. The initial position of the cart is +2 m and the initial velocity of the cart is 0.5 m/s 2. Sketch the situation - The sketch shows: The direction of the motion of the object. The negative and positive direction The initial velocity of the object. The initial position of the object. The initial clock reading of the object + vxi = +0.3 m/s dxi = +2 m ti = 0 s 3. Motion diagram V= - v1 v2 v3 + v4 The motion diagrams shows: The object represented as a particle. The position of the object each second. V arrows V arrows 4. Data table POSITION dx [m] CLOCK READING t [s] +2.0 m 0s +2.4 m 1s +3.0 m 2s +3.8 m 3s +4.8 m 4s +6.0 m 5s The data table shows: The position of the object each clock reading. 5. Position vs. clock reading graph The position vs clock reading graph shows: POSITION POSITION VS. CLOCK READING 16 14 12 10 8 6 4 2 0 Shape: Parabola Initial position = Intercept (the point where the parabola crosses the vertical axis) Mathematical model: dXF = ( ½ · aX · t2 ) + ( vXi · t ) + dXi 0 1 2 3 CLOCK READING 4 5 6. Velocity vs. clock reading graph The velocity vs clock reading graph shows: VELOCITY VELOCITY VS. CLOCK READING 6 5 4 3 2 1 0 Shape: Diagonal straight line. Slope = acceleration Mathematical model: vXF = ( aX · t ) + vXi 0 2 4 6 CLOCK READING Finding the slope: Remember, the slope of the graph is the acceleration of the object. Choose any two points on the graph Mathematical model: ax = Substitutions aX = Answer with units 4.5 vxf - vxi tf - ti m m - 1.5 s s 4s - 1 s aX = 1 m = s 3s 3 m s2 The acceleration of the object is aX = 1 m s2 7. Acceleration vs. clock reading graph The acceleration vs clock reading graph shows: ACCELERATION ACCELERATION VS. CLOCK READING 1.2 1 0.8 0.6 0.4 0.2 0 Shape: Horizontal straight line. Mathematical model: aX = constant 0 2 4 CLOCK READING 6 8. Mathematical models The mathematical model for the position vs. clock reading graph is dXF = ( ½ · aX · t2 ) + ( vXi · t ) + dXi, we substitute the acceleration (aX), the initial velocity (vxi), and the initial position (dxi): dxf = ( 1 m · 1 2 ·t2 )+ (0.5 m/s ·t) + 2 m 2 s The mathematical model for the velocity vs. clock reading graph is vXF = ( aX · t ) + vXi, we substitute the acceleration (aX), and the initial velocity (vxi). vxF = (· 1 This mathematical model describes the position of the object at any given clock reading m · t )+ 0.5 m/s s2 This mathematical model describes the velocity of the object at any given clock reading The mathematical model for the acceleration vs. clock reading graph is ax = constant, we substitute the acceleration (ax) aX = constant This mathematical model describes the acceleration of the object at any given clock reading PREDICTIONS Mathematical models are used to make predictions. In motion with constant velocity we can make predictions such as HOW FAR (displacement), HOW FAST (Velocity), HOW LONG (time) To make it easy we transform the mathematical model: vx = dxf - dxi tf - ti , into a triangle: vxf - vxi ACCELERATION vx ax t HOW LONG HOW LONG t = v XF - v Xi aX Use this mathematical model to make predictions about the time it takes an object to reach certain velocity. VELOCITY AFTER ANY DISPLACEMENT Use this mathematical model to make predictions: When t is not given. When you need to find the velocity of the object after any displacement. Do not forget to take the square root at the end of your calculations ( vXF )2 = ( vXi )2 + ( 2 · aX · dX ) DISPLACEMENT (dx) There are two ways to find the displacement of an object. 1. Using the Position vs. Clock reading graph: The displacement is the change in position POSITION VS. CLOCK READING 1. With an arrow identify the initial position (dxi). POSITION 2. With an arrow identify the Final position (dxF). 16 14 dxF =12 10 8 dxF =564 2 0 3. With an arrow identify the displacement; always from (dxi) to (dxF). dx 0 4. Find the displacement with a mathematical model: dx = dXF - dXi 1 2 3 4 dx = 12 m – 5 m dx = 7 m 5 CLOCK READING 2. Using the velocity vs. Clock reading graph: The displacement is the area bounded between the axis and the diagonal straight line (the shape of a trapezoid). VELOCITY VS. CLOCK READING VELOCITY 6 1. With an arrow identify base 1 (vxi). 5 2. With an arrow identify base 2 (vxF). 4 3 3. With a horizontal arrow identify the change in clock reading (t). vxF 2 vxi 1 0 0 1 2 t 3 CLOCK READING 4 5 4. Find the displacement with a mathematical model: dx = ½ · ( vXF + vXi ) ·t dx = ½ · (4.5 m/s + 2.5 m/s) · 2 s dx = 7.0 m