## anonymous one year ago Guy riding a bike accelerates for 8 seconds. He covers 60 meters. Using the formula d=vt+1/2at^2 calculate his acceleration.

1. anonymous

Alright so basically plugging in the numbers you'd think right

2. anonymous

Look 60m=1/2(a)64s^2

3. anonymous

Units don't cancel out

4. anonymous

60m=32s^2a

5. anonymous

I mean wtf

6. anonymous

a=(m/s^2)

7. anonymous

8. anonymous

No way to derive units for acceleration with the given kinematics equation which holds true

9. anonymous

wtf

10. anonymous

Hey dude what do you think

11. anonymous

lol you stuck with this paradox too

12. anonymous

don't worry you are not the only one

13. anonymous

The only way to do this is to assume "a" doesn't bear any value.

14. anonymous

Nope.

15. anonymous

Merely a means to represent imaginary number which follows from the function of this equation

16. anonymous

$v=m/s$

17. anonymous

But from the principles of physics which assigns unit to every different letter this is a paradox

18. anonymous

eh?

19. anonymous

$d=m$ $a=m/s^2$

20. anonymous

Then the prospect of using a is misconceived because a in this equation doesn't assume the units m/s^2 No

21. anonymous

Instead I am left with the presupposition that a will have the units and must

22. anonymous

60m/32s^2=a

23. anonymous

total distance traveled=initial distance+(velocity)(time)+0.50(acceleration)(time)^2

24. anonymous

1.875m/s^2

25. anonymous

In this case the initial distance and initial velocity is ignored as the cyclist starts from the state of being at rest

26. anonymous

$m=m+(\frac{ m }{ s }*s)+0.50*\frac{ m }{ s^2 }*s^2$

27. anonymous

total m=m+m+0.50*m

28. anonymous

you are wrong

29. anonymous

you can't assume a to have (m/s^2) internally.

30. anonymous

the units of a are assigned subsequently from the calculation. Assigning such units without solid foundation only serves to confuse the user.

31. anonymous

acceleration is m/s^2

32. anonymous

Look 60m=32s^2a a=1.875m/s^2

33. anonymous

This is correct.

34. anonymous

But if you assume a=(s/m^2) that gets you nowhere.

35. anonymous

Try it for yourself with units to be accorded

36. anonymous

This is where the teaching of physics fails.

37. anonymous

Each representative is not to be assumed to have the units but rather guidelines for what they stand for and what the user should expect to get following the calculations.

38. anonymous

acceleration is velocity over time. Acceleration is how fast velocity is changing with time since velocity/time is acceleration. and velocity is m/s and time is s (m/s)*(1/s)=m/s^2

39. anonymous

By no means assign the units prior to calculation as it throws of the equation.

40. anonymous

But it's interesting.

41. anonymous

so we know acceleration is m/s^2 because acceleration is known from how fast velocity is changing. So they plug a into your equation and the units cancels.

42. anonymous

No

43. anonymous

a=v/t came first before d=d1+vt+1/2)at^2

44. anonymous

When a is imaginary you can't even assign m/s^2

45. anonymous

However when a is known you can naturally assign the units however "naturally" as in being circumstantial and not at the hands of calculator

46. anonymous

there is no imaginary. there is an estimated A.

47. anonymous

hypothetical A

48. anonymous

so that hypothetical a doesn't assume units in itself

49. anonymous

You know what I mean? So many amateurs like me are presupposing upon seeing the hypothetical a or v or even d that they have units in themselves but this assumption is totally off

50. anonymous

Yes, it can if velocity and time can too?

51. anonymous

and acceleration is derived from them

52. anonymous

This is quite contradictory from what physics teachers force you to assume.

53. anonymous

we use seconds for time and units for speed why not use acceleration by dividing the units out

54. anonymous

Nothing imaginary about that. It is not abstract like math.

55. anonymous

I think my best bet is whenever you see a letter that makes a reference to set of units you need to assume that they in themselves do not possess anything of the sort however it's a guideline instructing the user of that equation to assume that the final work will be concordant with the standard form.

56. anonymous

Yeah physics is very paradoxical sometimes but that is our passion right? XD

57. anonymous

Yes, but it is also inaccurate in a lot of ways.

58. anonymous

PV=nRT for ideal gas law n does not make sense for me also.

59. anonymous

that is my opinion on some part of physics, but I think the acceleration actually does make sense.

60. anonymous

Yeah. Sometimes physics ignores the principles of math and heavily relies on intuitive reasoning which often fails. But this is the reality.

61. anonymous

The prospect of entirely relying on single solution upon a problem of kind shows it all. Unfortunate but yet convenient.

62. anonymous

Yep, that is the difference between experimental and theoretical physicists. They try to prove each other wrong. Well I got to sleep now. See you another day.

63. anonymous

Yeah Interesting talk my friend;) Good to see you.