compagnia04

2021-12-18

Given the velocity v(t) of an object moving along a ccordinate line at time t, find the object's displacement $s\left(t\right)$ moving along the coordinate line. The objects initial displacement is zero
$v\left(t\right)=2{t}^{h}{\left({t}^{k}-2\right)}^{4}$

levurdondishav4

s(t) is nothing but the integration of v(t) over time.
That is,
$s\left(t\right)=\int v\left(t\right)dt$
And intial displacement is given to be 0, which implies, $s\left(0\right)=0$
On taking $h=2$ and $k=3$ we have
$s\left(t\right)=\int 2{t}^{2}{\left({t}^{3}-2\right)}^{4}dt$
Now we will be solving the above integral :
$s\left(t\right)=\int 2{t}^{2}{\left({t}^{3}-2\right)}^{4}dt$
$⇒s\left(t\right)=2\int {t}^{2}{\left({t}^{3}-2\right)}^{4}dt$
We will take below substitution:
${t}^{3}-2=u⇒3{t}^{2}dt=du⇒{t}^{2}dt=\frac{du}{3}$
${t}^{3}-2=u⇒3{t}^{2}dt=du⇒{t}^{2}dt=\frac{du}{3}$
The integral becomes:
$s\left(t\right)=\frac{23}{\int }{u}^{3}du$
$⇒s\left(t\right)=2\int {t}^{2}{\left({t}^{3}-2\right)}^{4}dt$
$⇒s\left(t\right)=\frac{2{u}^{5}}{15}+C$
Putting back the value of u, we get
$s\left(t\right)=\frac{2{\left({t}^{3}-2\right)}^{5}}{15}+C$
where C is a constant
Now s(0)=0
This implies,
$s\left(0\right)=\frac{2{\left({0}^{2}-2\right)}^{5}}{15}+C=0$
$⇒0=\frac{{2.2}^{2}}{15}+C$
$⇒0=\frac{{2}^{6}}{15}+C$
$⇒-\frac{{2}^{6}}{15}=C$
$⇒-\frac{64}{15}=C$
Therefore, finally, we get
$s\left(t\right)=\frac{2{\left({t}^{3}-2\right)}^{5}}{15}$

abonirali59