Practice Problem Mr. Ant, standing in an elevator, moves up 40 m with the elevator. He then gets out of the elevator and walks along a straight hallway for 3 minutes at a speed 10 meters/minute. What is the magnitude of Mr. Ant’s net displacement?
What is the difference between… Speed Average velocity Velocity Instantaneous velocity Acceleration
Practice Problem Which of the following cars has a westward acceleration? A car traveling eastward and speeding up. A car traveling westward and slowing down. A car traveling westward at constant speed. A car traveling eastward and slowing down.
Solving kinematics problems Remember the fundamental kinematics equations. Know which quantities each equation has, and doesn’t have. Fundamental kinematics equations only apply for constant velocity. Use the table of variables – unique to 5S, and very useful!
Kinematics Problem Solving Step 1: Write out all five variables in a table. Fill in known values, put “?” next to unknown values. Step 2: If 3 or more known values, continue to Step 3. Otherwise, check your work. Step 3: Choose the equation that has all of the knowns and the desired unknown. Solve. Step 4: Put the correct units on your answer.
Freefall Acceleration due to gravity is a = -10 m/s 2 when gravity is the only force on the object. The horizontal motion of an object will NOT affect its fall – time to fall depends only on the vertical position/motion.
Example Problem What is the watermelon’s velocity when it hits the ground?
Practice Problem What is the watermelon’s velocity when it hits the ground? downward
Projectile Motion Break the velocity into vector components. We can work with the vector components separately because they are independent of each other. The y-component of velocity, v y, equals v(sin θ). The x-component of velocity, v x, equals v(cos θ). Horizontal velocity is constant. Vertical acceleration is g, directed downward. Time is the variable that is the same for x and y.
Motion Graphs The slope of a position–time graph at any point is the velocity of the object at that point in time. The slope of a velocity–time graph at any point is the acceleration of the object at that point in time. The area under a velocity–time graph between two times is the displacement of the object during that time interval.
Practice Problem The following data shows the positions of a car at times 0, 2, 4, 6, 8 and 10 s. Match the list of positions with the graph to which it corresponds. 0.0, 32.5, 65.0, 97.5, 130.0, 167.5 m. 0.0, 8.0, 32.0, 72.0, 120.0, 168.0 m. 0.0, 55.0, 120.0, 168.0, 168.0, 168.0 m.
Practice Problem The graph shows the speed of a car traveling in a straight line as a function of time. The value of V c is 3.80 m/s and the value of V d is 5.30 m/s. Calculate the distance traveled by the car from a time of 1.20 to 5.20 seconds.
Super Challenge Problem A student walks into an elevator at rest on the bottom floor of a building with 4 stories. The elevator then accelerates upward with an unknown constant acceleration a during a time interval Δt1 = t, moves at a constant velocity for Δt2 = 6t, and then decelerates at the same magnitude a for time interval Δt3 = t. If the elevator rises a total of h meters, what is the magnitude of the acceleration a given time t and height h?