physics ramp question

[rofl]

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a block goes down a frictionless ramp in 3 seconds. what is the ratio of the distance is travels in the 1st second compared to the distance is travels in the 3rd second?

mathematical solution, if possible. thanks!

 

[stsa84]

Edit: MT Headed is correct, I'll rework it when I have time if nobody else answers it.

 

[MT Headed]

At t = 3s: Δx = ½ a(9)

Read the question carefully. You solved for a Δx when Δt is 3 (i.e. tf=3, ti=0). That is not what the question is asking.

 

[Isoprop]

Use d = vi + 0.5at^2.
Assume initial velocity is zero (starts from rest).

After the first second: the block goes down a distance 0.5a
After the second second: block goes down a distance of 2a
After the third second: block goes down a distance of (9/2)a

The distance it goes during the third second is the total distance traveled, (9/2)a, minus the distance traveled after the 2nd second, 2a.

That is (5/2)a.

So the ratio is (1/2)a to (5/2)a which is 1:5

 

[SephirothXR]

The ramp doesn't even matter for this question. They didn't give you an angle. The angle could be 1 degree. Or it could be 89 degrees. In that case, the ratio must be the same for all degrees, which would include 89.999999999999 degrees. You assume your own degree. Mine is a 90 degree ramp =P. Just use distance = .5gt^2. Plug in 1, 2, and 3. Subtract (3-2) from (1-0) to get your answer.

I got 1:5

 

[neurodoc]

The ramp doesn't even matter for this question. They didn't give you an angle. The angle could be 1 degree. Or it could be 89 degrees. In that case, the ratio must be the same for all degrees, which would include 89.999999999999 degrees. You assume your own degree. Mine is a 90 degree ramp =P. Just use distance = .5gt^2. Plug in 1, 2, and 3. Subtract (3-2) from (1-0) to get your answer.

I got 1:5

Your reasoning is good. The inclined plane is irrelevant to the answer, since the question asks for the ratio of the distance (and not the actual distances travelled). The problem would be the same if the object were just dropped in the (frictionless) air and hit the ground after 3 seconds. In the 3rd second it travelled a distance of 1/2g3^2 (4.5g "units of distance"). In the first second it travelled 1/2g1^2 (1/2g "units of distance"). The ratio of these two distances is 9. Does this amke sense?🙂

 

[Arctangent]

Since the force pushing the block down the ramp is constant (and there is no friction), you have uniform acceleration, so like everyone was saying, the ramp part doesn't matter. They could have pitched you this same question saying you were on another planet with weaker acceleration due to gravity, and the calculations would be the same.

One way to look at this problem is if you are a bit familiar with integrals and how it relates to motion, you know that the parent function of uniformly accelerated motion is the parabola. Using x = a*t^2, where the a is some constant that represents acceleration, x is the displacement, and t is the time in s, you can figure out they are asking for the change in x, between 0s and 1s compared to 2s and 3s. By knowing a few easy squares, you can figure out that:
Displacement between 0s and 1s: a*1^2 - a*0^2 = 1*a
Displacement between 2s and 3s: a*3^2 - a*2^2 = 5*a
Thus, the ratio is 5:1.

This is essentially the same process as using the actual formula x = (1/2)a*t^2, assuming some unknown but constant acceleration a, and solving for the distance traveled at time points 0s, 1s, 2s, and 3s. The difference between the two formulas is just that in the first, the (1/2) is integrated into the constant, as it is in fact a constant value.