The equation is: $S = ut+ \frac{at^2}{2}$
How can I rearrange second equation of motion to find t?
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algebra-precalculus
classical-mechanics
2 Answers
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HINT: $$s=ut+\frac{1}{2}at^2$$ $$\implies at^2+2ut-2s=0$$ $$\implies t=\frac{-2u\pm\sqrt{4u^2+8as}}{2a}$$ $$\implies t=\frac{-u\pm\sqrt{u^2+2as}}{a}$$
Leave the negative one, as in classical macehenics, time does not assume negative value.
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Write $$s=ut+\frac{1}{2}at^{2}$$ Thus $$0=at^{2}+2ut-2s$$ Assume $a \neq 0$, whence $$0=t^2+\left(\frac{2u}{a}\right)t-2\left(\frac{s}{a}\right)$$ Complete the square on the terms in $t$ $$0=\left(t+\frac{u}{a} \right)^{2}-\frac{u^{2}}{a^{2}}-\frac{2s}{a}$$ hence $$\left( t+\frac{u}{a}\right)^{2}=\frac{1}{a^{2}}\left(u^{2}+2as \right)$$ From which $$t+\frac{u}{a}=\pm \frac{1}{a}\sqrt{u^{2}+2as}$$ And thus $$t=-\frac{u}{a}\pm \frac{1}{a}\sqrt{u^{2}+2as}$$ I'll leave it as an exercise as to whether both $\pm$ cases are required.