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Here, by tangential acceleration, I mean the component of acceleration along the velocity vector. What do you get when you integrate tangential acceleration with respect to time? What does the '$v$' that we get represent? I've heard that tangential acceleration is the rate of change of speed or $d|\vec{v}|/dt$ but speed is a scalar, while acceleration is a vector.

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Note that
$$2 \vec{v} \cdot \vec{a}~=~ 2 \vec{v} \cdot \frac{d\vec{v}}{dt}~=~\frac{d\vec{v}^2}{dt}~=~ \frac{d|\vec{v}|^2}{dt}~=~2|\vec{v}|\frac{d|\vec{v}|}{dt} , \tag{1} $$

so the tangential acceleration, i.e. the component of acceleration along the velocity vector, is $$a_{\parallel}~:=~\frac{\vec{v}}{|\vec{v}|} \cdot \vec{a}~\stackrel{(1)}{=}~\frac{d|\vec{v}|}{dt}.\tag{2}$$ It is an $\mathbb{R}$-valued scalar, which could be negative. Conversely, $$ |\vec{v}|~\stackrel{(2)}{=}~\int\! \mathrm{d}t~a_{\parallel}.\tag{3}$$

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  • $\begingroup$ But speed cannot be negative. Consider a case where we have to describe a particle with tangential retardation( which we have as a function of time/dist)? In this case the tangential acceleration would be negative since it's direction is opposite of velocity. Then on integrating we would get v(which should be speed). But and at some value of t, 'v' would become negative. But if 'v' is speed, that shouldn't happen. $\endgroup$ – xasthor Feb 8 '17 at 11:36
  • $\begingroup$ Yes, speed is non-negative, but change in speed could be negative. $\endgroup$ – Qmechanic Feb 8 '17 at 12:11
  • $\begingroup$ Yes, but for e.g consider a case where tangential acceleration $= -t$. Now $dv/dt = -t$ where $v$ is speed. On integrating, we get $v = v_0 - t^2/2$. Now if $t > \sqrt{2v_0}$, then $v$, which is speed, becomes negative. $\endgroup$ – xasthor Feb 8 '17 at 12:22
  • $\begingroup$ Right, so you have proven that $a_{\parallel}=-t$ is not realizable on the whole time axis $t\in\mathbb{R}$. $\endgroup$ – Qmechanic Feb 8 '17 at 12:41
  • $\begingroup$ What exactly do you mean by that statement? Since retardation here only depends on time, what is stopping $v = v_0 - t^2/2$ from becoming negative after $t > \sqrt{2v_0}$ $\endgroup$ – xasthor Feb 8 '17 at 13:21
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If you integrate the magnitude of the tangential acceleration vector, you should obtain the speed: $$ \int \mathrm dt \, |\vec a_\parallel(t)| = |\vec v(t)| \,. $$ This way, both sides of the equation have a scalar.

Integrating just the tangential component of the acceleration vector will probably be something that has no sensible interpretation. The integral of the full acceleration vector will be the velocity $\vec v$, though.

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  • $\begingroup$ But what about a case where we have to describe a particle with tangential retardation( which we have as a function of time/dist)? In this case the tangential acceleration would be negative since it's direction is opposite of velocity. Then on integrating we would get v. But and at some value of t, 'v' would become negative. But if 'v' is speed, that shouldn't happen. $\endgroup$ – xasthor Feb 8 '17 at 11:03
  • $\begingroup$ Valid point; one has to be careful when to take the absolute value. I'd just do all calculations with full vectors and then take the absolute value at the very end when I want to present a speed. Physics uses the velocity, speed is a rather unhelpful quantity, usually. $\endgroup$ – Martin Ueding Feb 8 '17 at 13:49

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