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

1. 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?

2. UnkleRhaukus

$$\theta$$|dw:1440728525844:dw|

3. UnkleRhaukus

|dw:1440728672933:dw|

4. UnkleRhaukus

|dw:1440728734123:dw|

5. 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?

6. idku

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

7. UnkleRhaukus

$W_y+N=0$

8. UnkleRhaukus

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

9. 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)

10. UnkleRhaukus

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

11. idku

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

12. UnkleRhaukus

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

13. 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.

14. idku

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

15. 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

16. 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.

17. UnkleRhaukus

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

18. 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.

19. UnkleRhaukus

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

20. UnkleRhaukus

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

21. UnkleRhaukus

so $g\sin\theta/2 =1$ $g\sin\theta =2$ $\therefore a= 2$

22. UnkleRhaukus

$\implies\qquad x(t) = t^2$

23. UnkleRhaukus

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

24. 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....

25. 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.

26. idku

I am just poor at maths, sorry-:(

27. UnkleRhaukus

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

28. idku

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

29. UnkleRhaukus

Are all the necessary understandings clear now?

30. 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])

31. 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$2=10\sin\theta\qquad\implies \theta = \arcsin(1/5) \approx 11.5°$