I need help with basic question. (I need the reasoning not the answer)

- idku

I need help with basic question. (I need the reasoning not the answer)

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- katieb

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- idku

At t=0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. You notice it moves 1 foot between t=0 seconds and t = 1 second.
1) How far does it move between t = 1 second and t = 2 seconds?

- UnkleRhaukus

\(\theta\)|dw:1440728525844:dw|

- UnkleRhaukus

|dw:1440728672933:dw|

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- UnkleRhaukus

|dw:1440728734123:dw|

- idku

I haven't ever done anything like that. Couldn't we relate to the fact that acceleration is the second derivative of the displacement?

- idku

MY teacher hasn't gone through anything of this sort in class. I am really sorry that you typed all of that.

- UnkleRhaukus

\[W_y+N=0\]

- UnkleRhaukus

\[\sum F = W+N\\
m\ddot x= mg \sin\theta \]

- idku

d=\(\dfrac{1}{2}\)\(\times\)a\(\times\)t\(^2\)
this is the relation to an acceleration in terms of a, am I correct? (where t is time)

- UnkleRhaukus

yeah,
and we have
\[a = \ddot x = g\sin \theta\]

- idku

I really have no clue about the x with two dots on top-:(

- UnkleRhaukus

the dots are time derivatives \[\ddot x = \frac{\mathrm d^2x}{\mathrm dt^2}\]

- idku

x=\(\dfrac{1}{2}\)at\(^2\)
when x=1 and t=1 (as in our case)
1=\(\dfrac{1}{2}\)a(1)\(^2\) --> a=2
Am I going in the correct direction? I am just trying to apply that one fact about the acceleration that it is 2nd derivative of positon with respect to time.

- idku

(x=1 and t=1, because it goes a distance of 1 foot from t=0 to t=1)

- idku

Now I will find the distance the ball travelled from t=0 to t=2
(I know t=2 because I am looking for displacement after 2 second, and I know a=2 because that is my acceleration)
x=\(\dfrac{1}{2}\)(2)(2)\(^2\)
x=4

- idku

So I found that from t=0 to t=1 it travelled 1 foot, and from t=0 to t=2 it travelled 4 feet.
So, between t=1 and t=2 it travelled 4-1=3 feet.

- UnkleRhaukus

We found the acceleration to be (a constant) \[a(t) = g\sin\theta\]
Integrating this twice, with respect to time:
\[x(t) = \iint a(t)\,\mathrm dt\,\mathrm dt\\
= g\sin theta\iint \,\mathrm dt\,\mathrm dt\\
= g\sin\theta t^2/2+c\]

- idku

I am stupid, indeed. I like the more elementary way-:(
I don't understand what you accomplished, and for you it is obvious as daylight.

- UnkleRhaukus

setting the origin to be at x(t=0) = 0
implies \[ x(t) = g\sin\theta\cdot\frac {t^2}2\]

- UnkleRhaukus

\[x(t=1)=g\sin\theta \cdot(1)^2/2=1[\text{ft}]\]

- UnkleRhaukus

so \[g\sin\theta/2 =1\]
\[g\sin\theta =2\]
\[\therefore a= 2\]

- UnkleRhaukus

\[\implies\qquad x(t) = t^2\]

- UnkleRhaukus

yes your right, it it moves 3 [ft] from t=1 to t=2

- idku

I, for my huge math inability, when I left a paper from calculus I my relatives were saying that I am a genius, even though it is fairly simple. Same thing is when you proposed your approach with is simple to you, but to me....

- idku

Thanks for your input though.... some time later, I will probably recall you posted something like this and get back to it when my knowledge will be efficient for that.

- idku

I am just poor at maths, sorry-:(

- UnkleRhaukus

i was looking at the physics, (it wasn't really necessary, but you asked for reasoning )

- idku

Well, what I said contained reasoning. Or didn't it? (My head is going crazy after a wild 16h day. Apologize again)

- UnkleRhaukus

Are all the necessary understandings clear now?

- UnkleRhaukus

You method was simple
acceleration is constant, so the position changes as x(t) = at^2/2
you used the point (t=1[s], x=1[ft]) to find the value of a=2
and then found the value at t=2[s] with x(t) = at^2/2 =t^2
the distance traveled between t=1&2 was the difference x(t=1[s]) - x(t=1[s])

- UnkleRhaukus

I can find the angle \(\theta\),
from the diagram, where the slope is at an angle \(\theta\)
\[a = g\sin theta\]
gravity \(g\approx10\)
and we found acceleration to be \(a=2\)
so\[2=10\sin\theta\qquad\implies \theta = \arcsin(1/5) \approx 11.5°\]

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