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idku
 one year ago
I need help with basic question. (I need the reasoning not the answer)
idku
 one year ago
I need help with basic question. (I need the reasoning not the answer)

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idku
 one year ago
Best ResponseYou've already chosen the best response.1At 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
 one year ago
Best ResponseYou've already chosen the best response.2\(\theta\)dw:1440728525844:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440728672933:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440728734123:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.1I haven't ever done anything like that. Couldn't we relate to the fact that acceleration is the second derivative of the displacement?

idku
 one year ago
Best ResponseYou've already chosen the best response.1MY teacher hasn't gone through anything of this sort in class. I am really sorry that you typed all of that.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2\[\sum F = W+N\\ m\ddot x= mg \sin\theta \]

idku
 one year ago
Best ResponseYou've already chosen the best response.1d=\(\dfrac{1}{2}\)\(\times\)a\(\times\)t\(^2\) this is the relation to an acceleration in terms of a, am I correct? (where t is time)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2yeah, and we have \[a = \ddot x = g\sin \theta\]

idku
 one year ago
Best ResponseYou've already chosen the best response.1I really have no clue about the x with two dots on top:(

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2the dots are time derivatives \[\ddot x = \frac{\mathrm d^2x}{\mathrm dt^2}\]

idku
 one year ago
Best ResponseYou've already chosen the best response.1x=\(\dfrac{1}{2}\)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
 one year ago
Best ResponseYou've already chosen the best response.1(x=1 and t=1, because it goes a distance of 1 foot from t=0 to t=1)

idku
 one year ago
Best ResponseYou've already chosen the best response.1Now 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
 one year ago
Best ResponseYou've already chosen the best response.1So 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 41=3 feet.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2We 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
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.2setting the origin to be at x(t=0) = 0 implies \[ x(t) = g\sin\theta\cdot\frac {t^2}2\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2\[x(t=1)=g\sin\theta \cdot(1)^2/2=1[\text{ft}]\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2so \[g\sin\theta/2 =1\] \[g\sin\theta =2\] \[\therefore a= 2\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2\[\implies\qquad x(t) = t^2\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2yes your right, it it moves 3 [ft] from t=1 to t=2

idku
 one year ago
Best ResponseYou've already chosen the best response.1I, 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
 one year ago
Best ResponseYou've already chosen the best response.1Thanks 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
 one year ago
Best ResponseYou've already chosen the best response.1I am just poor at maths, sorry:(

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2i was looking at the physics, (it wasn't really necessary, but you asked for reasoning )

idku
 one year ago
Best ResponseYou've already chosen the best response.1Well, what I said contained reasoning. Or didn't it? (My head is going crazy after a wild 16h day. Apologize again)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2Are all the necessary understandings clear now?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2You 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
 one year ago
Best ResponseYou've already chosen the best response.2I 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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