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How can I get v(t) out of this acceleration graph?

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  • $\begingroup$ Area under the graph. Note it’s zero for the middle interval, and negative in the third. $\endgroup$ – joseph h Mar 19 at 6:07
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$$a(t)=\frac{\mathrm{d}v(t)}{\mathrm{d}t}$$ $$\implies v(t)=\int a(t)\,\mathrm{d}t+C$$ If you would like to use explicit limits of integration, consider the particle at two times $t_1$ and $t_2$. Then, we can separate and integrate both sides: $$\int_{t_1}^{t_2}a(t)\,\mathrm{d}t=\int_{v(t_1)}^{v(t_2)}\,\mathrm{d}v(t)=v(t_2)-v(t_1)$$

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  • $\begingroup$ How do I apply the information I get from the graph to this equation? $\endgroup$ – Kali Mar 19 at 5:12
  • $\begingroup$ To start, use $a(t)=1$ for $0<t<3$. $\endgroup$ – G. Smith Mar 19 at 6:18

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