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# 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

Mayajyothi C J
126 Points
4 years ago
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x = t { (t-1) (t-2) }
x = t { t- 3t + 2 }
x = t+ 3t- 2t
Given v = 0
$\frac{\mathrm{d} x}{\mathrm{d} t}=0$
$\frac{\mathrm{d} (t^{3}-3t^{2}+2t)}{\mathrm{d} t}=0$
3t- 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 (t- 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.