Taking for example $v = \cos(t-1)$ from $t \in [0,1]$ and $v = e^{t-1}$ from $t \in (1,\infty)$ and $t \ge 0$. At $t = 1$, the function shifts from cosine to exponential, but remains continuous since $\cos 0 = 1$ and $e^0 = 1$. However acceleration is not the same for left hand side and right hand side. Is this possible?
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There is no problem with a discontinuity in the $a(t)$ graph. Actually it is a quite common situation in daily life.
Think for example of the following scenario:
A lamp is fixed to a cord hanging down from the ceiling.
Suddenly (at $t=t_0$) the cord breaks and the lamp begins falling down.
That means the acceleration of the lamp is
$$a(t)=\begin{cases} 0 &\text{for }t<t_0 \\ -g &\text{for }t>t_0 \end{cases}$$
And the velocity of the lamp is continuous with a kink at $t=t_0$: $$v(t)=\begin{cases} 0 &\text{for }t\le t_0 \\ -g(t-t_0) &\text{for }t\ge t_0 \end{cases}$$
answered Jun 20, 2021 at 9:10