David S. answered 11/10/19
Experienced AP/College Physics Teacher and Tutor
a) This depends on what is meant by "motion diagram". It could mean a dot diagram (showing the distance between positions for constant time intervals of a series of dots) or a position versus time graph. For the former, think of what would be happening to the distance between the dots (that represent equal time intervals) while it was moving at a constant speed versus when it was accelerating. For the latter, what is happening to the slope of the position-time graph while its speed is constant versus when it's accelerating. (Remember that the slope of the position-time graph is the velocity.)
b) The average acceleration is the change in velocity (final - initial) divided by the time interval.
c&d) Both diagrams have the same forces, what changes is the magnitude of one of the horizontal forces. Think about what the horizontal acceleration is (and when the horizontal forces are/aren't balanced).
e) The tricky thing here is that in order for the acceleration to double, the NET force has to double. Does that mean the pulling force doubles? Careful.
David S.
I believe that a student learns best when they are led to think through the solution rather than just given the answer (and the policy at Wyzant supports NOT just solving problems for students). I didn't "call the simple problem tricky". I referred to a part of the problem being "tricky" in that I've worked with enough students to know the most common mistake they make. And, if you think you can give a better solution without "the bunch of general words" (whatever that means), please post it.12/03/21
Grigoriy S.
12/03/21