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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.
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright so basically plugging in the numbers you'd think right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Look 60m=1/2(a)64s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Units don't cancel out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is such a paradox

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No way to derive units for acceleration with the given kinematics equation which holds true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hey dude what do you think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol you stuck with this paradox too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0don't worry you are not the only one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The only way to do this is to assume "a" doesn't bear any value.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Merely a means to represent imaginary number which follows from the function of this equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But from the principles of physics which assigns unit to every different letter this is a paradox

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then the prospect of using a is misconceived because a in this equation doesn't assume the units m/s^2 No

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Instead I am left with the presupposition that a will have the units and must

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0total distance traveled=initial distance+(velocity)(time)+0.50(acceleration)(time)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In this case the initial distance and initial velocity is ignored as the cyclist starts from the state of being at rest

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[m=m+(\frac{ m }{ s }*s)+0.50*\frac{ m }{ s^2 }*s^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can't assume a to have (m/s^2) internally.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the units of a are assigned subsequently from the calculation. Assigning such units without solid foundation only serves to confuse the user.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0acceleration is m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Look 60m=32s^2a a=1.875m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But if you assume a=(s/m^2) that gets you nowhere.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Try it for yourself with units to be accorded

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is where the teaching of physics fails.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Each 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
 one year ago
Best ResponseYou've already chosen the best response.0acceleration 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
 one year ago
Best ResponseYou've already chosen the best response.0By no means assign the units prior to calculation as it throws of the equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But it's interesting.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0a=v/t came first before d=d1+vt+1/2)at^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When a is imaginary you can't even assign m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0However when a is known you can naturally assign the units however "naturally" as in being circumstantial and not at the hands of calculator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is no imaginary. there is an estimated A.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that hypothetical a doesn't assume units in itself

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, it can if velocity and time can too?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and acceleration is derived from them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is quite contradictory from what physics teachers force you to assume.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we use seconds for time and units for speed why not use acceleration by dividing the units out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nothing imaginary about that. It is not abstract like math.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah physics is very paradoxical sometimes but that is our passion right? XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, but it is also inaccurate in a lot of ways.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0PV=nRT for ideal gas law n does not make sense for me also.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is my opinion on some part of physics, but I think the acceleration actually does make sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. Sometimes physics ignores the principles of math and heavily relies on intuitive reasoning which often fails. But this is the reality.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The prospect of entirely relying on single solution upon a problem of kind shows it all. Unfortunate but yet convenient.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah Interesting talk my friend;) Good to see you.
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