Question

Question #3d61e

Answer 1

`34.6` `m`/`s`

Explanation:

Use the equation: `d_x=1/2(V_2+V_1)(t)`

plug in `4700` `m` for `d_x` (`4.7` `km` is `4,700` meters)

Then plug in `6.6m`/`s` for `V_1`

And plug in `228` `s` for `t` (`3.8` minutes is `228` seconds)

so your equation should look like this:

`4700` `m` = `1/2(V_2+6.6` `m`/`s`)(`228`)

Divide `4700` `m` by `228` `s`, which equals `20.6140350877`

then divide by 0.5 and get `41.2280701754` `m`/`s`,

then subtract `6.6` `m`/`s` from that to get `34.6280701754`, or `34.6` `m`/`s`

Answer 2

`V_"final" = 34.6" m/s"`

Explanation:

To find the acceleration, a, use the equation:

`d=V_"initial"t+1/2at^2`

where `V_"initial" = 6.6" m/s"`, `d = 4.7" km" = 4700" m"`, and `t = 3.8" min" = 228" s"`

`4700" m"=(6.6" m/s")(228" s")+1/2a(228" s")^2`

`1/2a(228" s")^2=(4700" m"-(6.6" m/s")(228" s"))`

`a=2(4700" m"-(6.6" m/s")(228" s"))/(228" s")^2`

`a = 0.12293" m/s"^2`

To find the final speed, `V_"final"`, use the equation:

`V_"final"=sqrt(V_"initial"^2 +2ad)`

`V_"final"=sqrt((6.6" m/s")^2 +2(0.12293" m/s"^2)(4700" m"))`

`V_"final" = 34.6" m/s"`