On the Moon the acceleration .

In summary, the acceleration on the moon is 1.7 m/s2 and it would take a feather about 1.0m to fall that distance in 1.0s.
  • #1
bramill
3
0
On the Moon the acceleration...

On the Moon the acceleration of freefall is 1.7 m s-2. How much time would it take for a featherto fall a distance of 1.0m from rest?

Im not bothered about the answer i just need to know what formula(s) to use
and how.

These are the formulas i was given:

v = u + at
x = ut + 1/2 at^2
v^2 = u^2 +2ax
Average velocity = Displacement / Time
Average Velocity = (v+u)/2

u= initial velocity
a= acceleration
x= displacement
v= final velocity
t= time
 
Last edited:
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  • #2
bramill said:
On the Moon the acceleration of freefall is 1.7 m s-2. How much time would it take for a featherto fall a distance of 1.0m from rest?

Im not bothered about the answer i just need to know what formula(s) to use
and how.

These are the formulas i was given:

v = u + at
x = ut + 1/2 at^2
v^2 = u^2 +2ax
Average velocity = Displacement / Time
Average Velocity = (v+u)/2

u= initial velocity
a= acceleration
x= displacement
v= final velocity
t= time

But before you can use the formulae, you must understand them, no ?

What do the formulae mean to you ?

What values are you given ? What values can you infer, and what do you need to get ?

There's no point in just pointing out which formula to plug in values into to churn out the answer, that would be like just giving you the answer. You wouldn't learn much from that, I hope you understand.
 
  • #3
Well you can eliminate the equations that do not give you the information you need.

For example. V = U + at

You know what U is (0 m/s) and you know what a is but you don't know V or t.

So this can't be the equation you need to you use.

What about [tex] V^2 = U^2 + 2as [/tex]

Well, once again we know U and a and for that matter s. So we could use this equation to find V and then plug that into the first equation.

Or Being most sensible you could try

[tex] x = ut + \frac{1}{2} at^2 [/tex]

With these questions often the best way to find the answer is to start seeing which ones will give you useful information if you start plugging numbers in.
 
Last edited:
  • #4
i got an answer of 0.83 s to 2 d.p. can you tell me if this is correct please
 
  • #5
bramill said:
i got an answer of 0.83 s to 2 d.p. can you tell me if this is correct please

How did you get that (the working) ?
 
  • #6
the answer is 1.08 i just worked it out properly
 

Related to On the Moon the acceleration .

What is the acceleration on the Moon?

The acceleration on the Moon is approximately 1.62 meters per second squared, which is about one-sixth of the Earth's acceleration due to gravity.

How does the acceleration on the Moon compare to Earth's acceleration?

The acceleration on the Moon is about one-sixth of the Earth's acceleration due to gravity. This is because the Moon has less mass and a weaker gravitational pull compared to Earth.

Why does the Moon have a weaker acceleration compared to Earth?

The Moon has a weaker acceleration due to its smaller mass and weaker gravitational pull. This is because the Moon has a much smaller radius and less mass compared to Earth.

How does the Moon's lack of atmosphere affect its acceleration?

The Moon's lack of atmosphere has no direct effect on its acceleration. However, the lack of air resistance on the Moon's surface allows objects to fall at a constant acceleration, whereas on Earth, air resistance can affect the acceleration of falling objects.

Can the acceleration on the Moon change?

Yes, the acceleration on the Moon can change depending on the mass and distance of other objects in its vicinity. For example, the acceleration can increase or decrease if the Moon is in the gravitational pull of a larger object, such as a planet.

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