Math, asked by Anonymous, 1 month ago

pleas-e ans-wer hurry up​

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Answered by senboni123456
1

Step-by-step explanation:

We have,

 \frac{ {d}^{2} x}{d {t}^{2} }  = 3 {t}^{2} - 6t + 8  \\

  \implies \:  \frac{d}{dt}  \bigg(\frac{ d x}{dt}  \bigg) = 3 {t}^{2} - 6t + 8  \\

  \implies \:  d  \bigg(\frac{ d x}{dt}  \bigg) = (3 {t}^{2} - 6t + 8)dt  \\

  \implies \:   \int \: d  \bigg(\frac{ d x}{dt}  \bigg) =  \int(3 {t}^{2} - 6t + 8)dt  \\

  \implies \:   \frac{ d x}{dt}   =  3  \frac{{t}^{3}}{3} - 6 \frac{ {t}^{2} }{2} + 8t   + c\\

  \implies \:   \frac{ d x}{dt}   =  {t}^{3}- 3{t}^{2}  + 8t   + c\\

Since, v=\frac{dx}{dt}\\

And, at t=0, v=0

So,

  \implies \:   (0)  =  {(0)}^{3}- 3{(0)}^{2}  + 8(0)  + c\\

  \implies \:   c = 0\\

Required equation becomes,

  \implies \:   \frac{ d x}{dt}   =  {t}^{3}- 3{t}^{2}  + 8t   \\

  \implies \:   d x   =  ({t}^{3}- 3{t}^{2}  + 8t )dt  \\

  \implies \:   \int d x   =   \int({t}^{3}- 3{t}^{2}  + 8t )dt  \\

  \implies \:    x   =    \frac{ {t}^{4}}{4}- 3 \frac{{t}^{3} }{3} + 8 \frac{ {t}^{2} }{2}  + k  \\

  \implies \:    x   =    \frac{ {t}^{4}}{4}- {t}^{3}  + 4 {t}^{2}  + k  \\

At t=0, x=0,

So,

  \implies \:    (0)  =    \frac{ {(0)}^{4}}{4}- {(0)}^{3}  + 4 {(0)}^{2}  + k  \\

  \implies \:    k = 0  \\

Now, required distance, w.r.t time is

  \implies \:    x   =    \frac{ {t}^{4}}{4}- {t}^{3}  + 4 {t}^{2} \\

So, distance covered in 2 seconds will be,

  \implies \:    x(2)   =    \frac{ {(2)}^{4}}{4}- {(2)}^{3}  + 4 {(2)}^{2} \\

  \implies \:    x(2)   =    \frac{ 16}{4}- 8  + 4 \times 4\\

  \implies \:    x(2)   =    4- 8  +16\\

  \implies \:    x(2)   =    - 4  +16\\

  \implies \:    x(2)   =    12\\

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