MHB Is the Curve a Line or a Point with Zero Acceleration?

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A curve in \(\mathbb{R}^3\) with zero acceleration must be either a straight line or a point. The acceleration vector \(a(t)\) being zero implies that the second derivatives of the position functions \(x(t)\), \(y(t)\), and \(z(t)\) are all zero. Consequently, the position functions can be expressed as linear functions of time. The discussion raises the question of whether cases need to be considered when the coefficients of these linear functions are zero. Ultimately, it is confirmed that such cases are necessary to determine if the curve is indeed a straight line or a point.
mathmari
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Hey! :o

Let $\overrightarrow{\sigma}$ a curve in $\mathbb{R}^3$ with zero acceleration. Show that $\overrightarrow{\sigma}$ is a straight line or a point.

Let $\overrightarrow{\sigma}(t)=(x(t),y(t),z(t))$ be the curve.

We have that $a(t)=\overrightarrow{\sigma}''(x''(t),y''(t),z''(t))=(0,0,0)$

So $x(t)=c_1t+c_2, y(t)=c_3t+c_4, z(t)=c_5t+c_6$

Do we have to take cases if $c_1=c_3=c_5=0$ or not to conclude that the curve is a straight line or a point?? (Wondering)
 
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Heya! (Mmm)

mathmari said:
Do we have to take cases if $c_1=c_3=c_5=0$ or not to conclude that the curve is a straight line or a point?? (Wondering)

Yep. (Nod)
 
I like Serena said:
Heya! (Mmm)
Yep. (Nod)
Great! Thanks! (Smile)
 

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