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

$v\left(t\right)=2{t}^{h}{({t}^{k}-2)}^{4}$

levurdondishav4

Beginner2021-12-19Added 38 answers

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

$s\left(t\right)=\int 2{t}^{2}{({t}^{3}-2)}^{4}dt$

Now we will be solving the above integral :

$s\left(t\right)=\int 2{t}^{2}{({t}^{3}-2)}^{4}dt$

$\Rightarrow s\left(t\right)=2\int {t}^{2}{({t}^{3}-2)}^{4}dt$

We will take below substitution:

$t}^{3}-2=u\Rightarrow 3{t}^{2}dt=du\Rightarrow {t}^{2}dt=\frac{du}{3$

$t}^{3}-2=u\Rightarrow 3{t}^{2}dt=du\Rightarrow {t}^{2}dt=\frac{du}{3$

The integral becomes:

$s\left(t\right)=\frac{23}{\int}{u}^{3}du$

$\Rightarrow s\left(t\right)=2\int {t}^{2}{({t}^{3}-2)}^{4}dt$

$\Rightarrow s\left(t\right)=\frac{2{u}^{5}}{15}+C$

Putting back the value of u, we get

$s\left(t\right)=\frac{2{({t}^{3}-2)}^{5}}{15}+C$

where C is a constant

Now s(0)=0

This implies,

$s\left(0\right)=\frac{2{({0}^{2}-2)}^{5}}{15}+C=0$

$\Rightarrow 0=\frac{{2.2}^{2}}{15}+C$

$\Rightarrow 0=\frac{{2}^{6}}{15}+C$

$\Rightarrow -\frac{{2}^{6}}{15}=C$

$\Rightarrow -\frac{64}{15}=C$

Therefore, finally, we get

$s\left(t\right)=\frac{2{({t}^{3}-2)}^{5}}{15}$

That is,

And intial displacement is given to be 0, which implies,

On taking

Now we will be solving the above integral :

We will take below substitution:

The integral becomes:

Putting back the value of u, we get

where C is a constant

Now s(0)=0

This implies,

Therefore, finally, we get

abonirali59

Beginner2021-12-20Added 35 answers

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