Mathematics: Find the position and velocity of an object moving along a straight line given acceleration, initial velovity, and initial position.a(t)=8e^-tv(0)=-8s(0)=10

Find the position and velocity of an object moving

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14.01.2022

Step-by-step explanation:

Given

$a(t)=8e^{-t}$

and v(0)=-8

s(0)=10

we can write a in terms of velocity

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$8e^{-t}=\frac{\mathrm{d} v}{\mathrm{d} t}$

dv=8e{-t}dt

$\int dv=\int 8e^{-t}dt$

$v=-8e^{-t}+c$

substituting v(0)=-8

we get c=16

$v=-8e^{-t}+16$

Now $v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$ds=\left ( -8e^{-t}+16 \right )dt$

$\int ds=\int \left ( -8e^{-t}+16 \right )dt$

$s=8e^{t}+16t+c_1$

Putting s(0)=10

c_1=2

$s=8e^{t}+16t+2$

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