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I was hoping somebody could help me with a problem I've been working on and
possibly point me in the right direction if I've made a mistake. The
problem reads as follows:
From the top of a cliff, a person uses a slingshot to fire a pebble
straight downward, which is the negative direction. The initial speed of
the pebble is 9.0 m/s (a) What is the acceleration (magnitude and
direction) of the pebble during the downward motion? Is the pebble
decelerating? Explain. (b) After .50 s, how far beneath the cliff top is
For part (a) I have the acceleration of the pebble is 9.8 m/s^2 downward
and that it is not decelerating because the vectors do not point in
For part (b) my math looks as follows:
x = (v)(t) + (1/2)(a)(t^2)
x = (9.0 m/s)(.50s) + (1/2)(9.8 m/s^2)((.50s)^2)
x = 4.5m + 1.2m = 5.7m
I have a fair bit of confidence in my answer for part (b), however it is
part (a) that's been troubling me since I'm not really sure what is meant
by "the negative direction." As far as I understand when gravity causes
ojbects to accelerate then it is +9.8m/s^s and when objects are
decelerating due to gravity then it is -9.8m/s^2. Since the object
accelerates all the way to the bottom of the cliff, I don't see why any
values in this problem should be negative. Can someone please explain this
"nebulus" <[Only registered users see links. ]> wrote in message
The acceleration is the same regardless of the speed or posiiton (which
depend on other factors anyway). The magnitude is 9.8 m/sec^2 and the
direction is [by definition!] "down."
Your convention is that "down" is the negative direction. By this
convention, the velocity is negative AND the acceleration is negative.
Since they both have the same sign in your coordinate system, the pebble is
not decelerating (which only occurs when they signs of velocity and
acceleration are opposed).
Initial velocity = -9.0 m/sec
Acceleration = -9.8 m/sec^2
Interval = 0.5 sec
Change in velocity = -9.8 m/sec^2 * 0.5 sec = -4.9 m/sec
Final velocity = -9.0 m/sec + -4.9 m/sec = -13.9 m/sec
Average velocity throughout interval = (-9.0 + -13.9)/2 = -22.9/2 = -11.45
Distance travelled during interval = -11.45 m/sec * 0.5 sec = -5.725 m
Whether it is +9.8 or -9.8 m/sec^2 depends on which direction you define as
"Deceleration" implies a comparison between the directions of the velocity
and acceleration vectors.
You said, and I quote: "straight downward, which is the negative direction".
This is one of those problems where some prof can be playing word games. If the
negative direction is "down" and the acceleration is "down" I'd wonder if he's
looking for deceleration. Yeah, it's nuts, but if he takes the opposite of
acceleration as deceleration, wll then. . .
Which makes me think the teacher somewhere might have minored in law. The
answer as to displacement is correct, but the word games makes me not sure
about the terminalogy.
nebulus <[Only registered users see links. ]> wrote in message news:<Xns93EF8219A534Ena@184.108.40.206>...
Whatever your answer to part (a) is, have your reasons ready to
justify it. If your teacher marks you down for it, challenge him on
it. I didn't get A's in school by keeping my mouth shut!
[Only registered users see links. ]emoove (tony) wrote in
news:[Only registered users see links. ]:
Actually, the problem comes straight from my textbook. I left off the
first sentence which referred to a section of the book that discussed
deceleration but it wasn't quite as clear as the explanation I got from
this newsgroup. The book we are using is Cutnell & Johnson Physics, 5th
Ed. (2001). I purchased a study guide to go with it that has the solutions
and methods of solving maybe 5% of the problems in the book, which means it
really doesn't do me much good for 95% of the material we cover. This
newsgroup has been (and I hope it will continue to be in the future) an
invaluable resource in helping me understand these concepts. I'll search
google... perhaps there is a Cutnell & Johnson Law book floating around
[Only registered users see links. ] (Double-A) wrote in
Always good advice! I'm seriously going for that 4.0 (so far, so good).