Question

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

Answer 1

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`