My students will take the Mastery Interactive assessments online. There are a small number of questions on the free body diagram on velocity and acceleration. I looked everywhere on the slides and lectures (as well as other sources) to guide my thinking so I can impart this information to my students. I wasn&#x27;t helped much. Can you please guide me on how I can teach on the FBD on velocity and acceleration especially when the object is in a free fall and coming back to the ground again. Thank you so much!
Joan, Free body diagrams are taught in the Dynamics unit - the presentation starts on slide 98 with free body diagrams, then the rest of the presentation shows how to do the problems. Several FBD problems are solved step by step, starting on slide 130. For free fall, you really don't need a free body diagram - at this level, we ignore friction, so the only force acting on an object in free fall is mg. So, acceleration equals -g (downward direction) and you use the three kinematics equations to solve that problem. John
I think i got it confused. There was this question on Masterly interactive with the vector diagrams asking what arrow direction matches objects in free fall in velocity and acceleration. Is it possible to remove that question or can you please guide me on how to teach that. I am thinking that the velocity vector should point downward when thrown upward, or should the vector dire turn be upward since the direction is upward?
The velocity vector points in the direction of the motion. So when the object is going up, velocity vector is up. Object at the top momentarily stops, and then returns to the earth - and then the velocity vector is pointing down. The object is slowing down the minute it is launched until it reaches the maximum height (where velocity is zero) and then continually increases its velocity until it returns to the launcher with the same speed it took off with.
The acceleration vector ALWAYS points down as the acceleration is due to gravity, which always points down.
John
In slide 156 where you describe the derivation for KE 3, it would be simpler to substitute the expression for t from KE1 in the equation X-Xo = Vavg t rather than substitute in KE 2. This derivation may be more easily graspable to the students and could be included in the slides.
X-Xo = (V+Vo)/2 x (V-Vo)/a etc....
Jacob, you're right - but we made the call to start with the first 2 kinematics equations to show their connection. X-Xo = Vavg t is simpler, but it's not part of the "big 3." John
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