Modern Physics> Can you please help me with this :) I hop...

Can you please help me with this :) I hope my handwriting is clear this time Thanks a lot for looking into this question sir\maam

Aastha , 8 Years ago

Grade 11

1 Answers

Mayajyothi C J

Last Activity: 8 Years ago

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x = t { (t-1) (t-2) }

x = t { t²- 3t + 2 }

x = t³+ 3t²- 2t

Given v = 0

$\frac{\mathrm{d} x}{\mathrm{d} t}=0$

$\frac{\mathrm{d} (t^{3}-3t^{2}+2t)}{\mathrm{d} t}=0$

3t²- 6t + 2 = 0

Solving you get

$t = \frac{1 - (3)^{\frac{1}{2}}}{3^{\frac{1}{2}}}$

and $t = \frac{- (3)^{\frac{1}{2}-1}}{3^{\frac{1}{2}}}$

Substitute this in x,

you will get x. try doing it that way, I hope you could get the answer. I’m not good in algebra to solve further.

To find acceleration,

$a = \frac{\partial^2 x}{\partial t^2}$

= d² (t³- 3t² + 2t)/dt²

= 6t - 6

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Substitute t in the question. Hope this was helpful. It is very difficult to do all the algebraic solution for me. This answer is not complete I know it. But this the fartest explaination I can give, if you have any doubt in this explaination ask me in this page itself if you open it as a new query it would be difficult for me to locate you. As the doubt by entering it as an answer in the same page.

Please approve ^_^