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Speed as a Function of Time 
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#1
Jul214, 02:17 PM

P: 101

The first time derivative of velocity is acceleration. Can we then conclude that the first time derivative of speed is the magnitude of acceleration? In the following example I will consider a one dimensional case, for the sake of argument. Suppose the velocity v of a particle as a function of time t is given by v(t) = t^2. The acceleration, a, as a function of t is therefore given by a(t) = 2t. And so the magnitude of acceleration (the absolute value, since we are dealing with a one dimensional case as I have previously stated) is a "piecewise defined function of t", namely 2t. That's observation A.
Now, let's go back to velocity. Since v(t) = v(t), we can then conclude that the derivative of speed as a function of time is given by d/dt(v(t)) = 2t; which, technically speaking, is not the same as a(t). So am I right in my conclusion that differentiating speed does not yield the magnitude of acceleration? 


#2
Jul214, 02:25 PM

P: 4,068




#3
Jul214, 02:30 PM

P: 101




#4
Jul214, 02:56 PM

P: 101

Speed as a Function of Time
To sum it all up:
Mathematically, df(x)/dx [itex]\neq[/itex] df(x)/dx. Ergo the first time derivative of speed is not the magnitude of acceleration, am I right? 


#5
Jul214, 02:57 PM

P: 4,068




#6
Jul214, 03:04 PM

P: 316

dv/dt in general does not equal the magnitude of acceleration.
generally v is a vector with several components and the derivative is also a vector with several components. I don't readily know if dv/dt = dv/dt is true or false, but I would assume not. There might be some special cases where it is true. 


#7
Jul414, 03:27 PM

P: 19

Aren't we certain that dv/dt =/= dv/dt?
v(t) = v(t) and the derivative of v(t) is a(t) Then the derivative of v(t) would also be a(t) a(t) is not a(t) Therefore, the derivative of v(t) is not a(t). So we know they aren't the same. Correct me if I have any mistakes. That seems to be the problem, I think MohammadRady97 is spot on. The magnitude of the acceleration is df(x)/dx but you, OP, tried to do df(x)/dx, which as MohammadRady said (and I've shown explicitly above) are not the same thing. 


#8
Jul414, 03:33 PM

P: 101




#9
Jul414, 11:12 PM

P: 316

I got bored. I think this is overkill. You're all welcome.
Say [itex]\vec{v}(t)=f(t)\hat{\textbf{i}}+g(t)\hat{\textbf{j}}[/itex] [itex]v(t)=\sqrt{[f(t)]^2+[g(t)]^2}[/itex] then [itex]\vec{a}(t) = \frac{dv}{dt}(t)=\frac{df}{dt}\hat{\textbf{i}}+\frac{dg}{dt}\hat{ \textbf{ j}}[/itex] where [itex]a(t)=\frac{dv}{dt}(t)=\sqrt{\left(\frac{df}{dt}\right)^2+\left( \frac{dg}{dt} \right)^2}[/itex] and [itex]\frac{dv}{dt}(t)=\frac{1}{2}\left( [f(t)]^2+[g(t)]^2\right)^{1/2}\left[2f(t)\frac{df}{dt}+2g(t)\frac{dg}{dt}\right][/itex] If you try and solve [itex]a(t) = \frac{dv}{dt}[/itex] you find that this is only true if [itex]\left(f(t)\frac{dg}{dt}\right)^2 + \left(g(t)\frac{df}{dt}\right)^2 = f(t)g(t)(\frac{df}{dt})(\frac{dg}{dt})[/itex] or [itex]\frac{f(t)}{g(t)}\frac{dg/dt}{df/dt}+\frac{g(t)}{f(t)}\frac{df/dt}{dg/dt} = 1[/itex] Therefore, the time derivative of v(t) is not a(t) unless that condition is met. Through a little guessing, the only solutions I've found are when either f or g = 0, or f and g are constants. I hope this clears everything up. 


#10
Jul514, 07:26 AM

P: 101




#11
Jul514, 01:39 PM

P: 316

yes, f(t) and g(t) are the x and y components of velocity. I was lazy some and used f and g, but they are the same thing. Recall that the derivative of speed you were asking about is [itex]\frac{d}{dt}\left(v(t)\right)[/itex], and the magnitude of acceleration is dv/dt, so that's why the answer is in components of the velocity.



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