anonymous
  • anonymous
Guy riding a bike accelerates for 8 seconds. He covers 60 meters. Using the formula d=vt+1/2at^2 calculate his acceleration.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Alright so basically plugging in the numbers you'd think right
anonymous
  • anonymous
Look 60m=1/2(a)64s^2
anonymous
  • anonymous
Units don't cancel out

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anonymous
  • anonymous
60m=32s^2a
anonymous
  • anonymous
I mean wtf
anonymous
  • anonymous
a=(m/s^2)
anonymous
  • anonymous
This is such a paradox
anonymous
  • anonymous
No way to derive units for acceleration with the given kinematics equation which holds true
anonymous
  • anonymous
wtf
anonymous
  • anonymous
Hey dude what do you think
anonymous
  • anonymous
lol you stuck with this paradox too
anonymous
  • anonymous
don't worry you are not the only one
anonymous
  • anonymous
The only way to do this is to assume "a" doesn't bear any value.
anonymous
  • anonymous
Nope.
anonymous
  • anonymous
Merely a means to represent imaginary number which follows from the function of this equation
anonymous
  • anonymous
\[v=m/s \]
anonymous
  • anonymous
But from the principles of physics which assigns unit to every different letter this is a paradox
anonymous
  • anonymous
eh?
anonymous
  • anonymous
\[d=m\] \[a=m/s^2\]
anonymous
  • anonymous
Then the prospect of using a is misconceived because a in this equation doesn't assume the units m/s^2 No
anonymous
  • anonymous
Instead I am left with the presupposition that a will have the units and must
anonymous
  • anonymous
60m/32s^2=a
anonymous
  • anonymous
total distance traveled=initial distance+(velocity)(time)+0.50(acceleration)(time)^2
anonymous
  • anonymous
1.875m/s^2
anonymous
  • anonymous
In this case the initial distance and initial velocity is ignored as the cyclist starts from the state of being at rest
anonymous
  • anonymous
\[m=m+(\frac{ m }{ s }*s)+0.50*\frac{ m }{ s^2 }*s^2\]
anonymous
  • anonymous
total m=m+m+0.50*m
anonymous
  • anonymous
you are wrong
anonymous
  • anonymous
you can't assume a to have (m/s^2) internally.
anonymous
  • anonymous
the units of a are assigned subsequently from the calculation. Assigning such units without solid foundation only serves to confuse the user.
anonymous
  • anonymous
acceleration is m/s^2
anonymous
  • anonymous
Look 60m=32s^2a a=1.875m/s^2
anonymous
  • anonymous
This is correct.
anonymous
  • anonymous
But if you assume a=(s/m^2) that gets you nowhere.
anonymous
  • anonymous
Try it for yourself with units to be accorded
anonymous
  • anonymous
This is where the teaching of physics fails.
anonymous
  • 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.
anonymous
  • 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
anonymous
  • anonymous
By no means assign the units prior to calculation as it throws of the equation.
anonymous
  • anonymous
But it's interesting.
anonymous
  • 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.
anonymous
  • anonymous
No
anonymous
  • anonymous
a=v/t came first before d=d1+vt+1/2)at^2
anonymous
  • anonymous
When a is imaginary you can't even assign m/s^2
anonymous
  • 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
anonymous
  • anonymous
there is no imaginary. there is an estimated A.
anonymous
  • anonymous
hypothetical A
anonymous
  • anonymous
so that hypothetical a doesn't assume units in itself
anonymous
  • 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
anonymous
  • anonymous
Yes, it can if velocity and time can too?
anonymous
  • anonymous
and acceleration is derived from them
anonymous
  • anonymous
This is quite contradictory from what physics teachers force you to assume.
anonymous
  • anonymous
we use seconds for time and units for speed why not use acceleration by dividing the units out
anonymous
  • anonymous
Nothing imaginary about that. It is not abstract like math.
anonymous
  • 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.
anonymous
  • anonymous
Yeah physics is very paradoxical sometimes but that is our passion right? XD
anonymous
  • anonymous
Yes, but it is also inaccurate in a lot of ways.
anonymous
  • anonymous
PV=nRT for ideal gas law n does not make sense for me also.
anonymous
  • anonymous
that is my opinion on some part of physics, but I think the acceleration actually does make sense.
anonymous
  • anonymous
Yeah. Sometimes physics ignores the principles of math and heavily relies on intuitive reasoning which often fails. But this is the reality.
anonymous
  • anonymous
The prospect of entirely relying on single solution upon a problem of kind shows it all. Unfortunate but yet convenient.
anonymous
  • 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.
anonymous
  • anonymous
Yeah Interesting talk my friend;) Good to see you.

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