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If a rock falls for one second, what is its average speed during that second?

a) Zero ft/s
b) 1 ft/s
c) 4 ft/s
d) 16 ft/s
e) 32 ft/s

2007-07-20 15:35:30 · 7 answers · asked by ? 6 in Science & Mathematics Physics

The answer is D. Although the speed of the rock at the end of the one-second interval is 32 ft/s, its beginning speed was zero--it was, dropped from a rest position. So the average speed could not be 32 ft/s, anymore than it could be zero. Since the accelerationduring the fall was constant, the average speed is simply 16 ft/s, midway between zero and 32 ft/s. We distinguish between instantaneous speed, the speed at a particular instant and the overall or average speed. Incidentally, since the stone has an average speed of 16 ft/s, it must fall a distance of 16 feet during that second.

Note: Be careful not to confuse "How Fast" and "How Far"--Speed and Distance travled are different. Even more different is the idea of "how Fast Does How Fast Change"--and that's Acceleration!

2007-07-21 07:40:25 · update #1

7 answers

The acceleration of gravity is approximately 32 ft/s^2. The velocity of a free-falling object after some amount of time is simply:
v = g*t
where g is gravity (32 ft/s^2) and t is the time it has been falling, in seconds.

After 1 second, the rock will be going at 32 ft/s. The average velocity is simply half of this, or 16 ft/s. So, d) is the correct choice.

2007-07-20 15:48:37 · answer #1 · answered by lithiumdeuteride 7 · 1 0

Acceleration (a) = 32 ft / sec^2
Time (t) = 1 sec

Distance = 1/2 at^2 or 1/2 (32 ft/sec^2) x 1sec^2 or
Distance = 16 ft

Average Speed = Distance / Time, or
Avg Spd = 16 ft / 1 sec, or
16 ft/sec, which is answer (d)

2007-07-20 15:47:16 · answer #2 · answered by Kevin S 7 · 1 0

Average speed = total distance / total time

total time is 1 s
tolal distance the rock falls in one second is:
y = .5(32)(1^2)
y = 16ft

so 16ft/1s = 16ft/s

the answer is D

2007-07-20 15:41:32 · answer #3 · answered by      7 · 1 0

ok im gunna ignore ur choices cuz there all in feet but last year in science i learned that everything falls at a speed of 9.8 m/s/s or m/s squared so w/e that would be in feet i guess so D?

2007-07-20 15:41:35 · answer #4 · answered by Anonymous · 1 0

You write the equation of action as m dv.dt = - mg - nu v right here nu is the frictional rigidity because of environment. g is the acceleration because of gravity. you stumble on the condition whilst dv/dt=0 this could provide the ordinary % the ordinary % is v = mg/nu This in many situations is 6 km/sec.

2016-11-10 00:40:27 · answer #5 · answered by ? 4 · 0 0

answer is e.
a=32ft/s^2
t=1s
initial velocity(u)=0
v=u+at
=0+32*1
=32ft/s
or in SI units

v=0+9.8*1
=9.8m/s

2007-07-21 04:24:33 · answer #6 · answered by Anonymous · 0 0

i say C, b/c i think the avg rate of speed over time is like 3.19 or something.

2007-07-20 15:38:29 · answer #7 · answered by jasx501 3 · 0 0

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