Question

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Answer 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,

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

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

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

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

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