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

Answer:

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 $s(0)=10$

$c_1=2$

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