# Question 3d61e

Oct 12, 2017

$34.6$ $m$/$s$

#### Explanation:

Use the equation: ${d}_{x} = \frac{1}{2} \left({V}_{2} + {V}_{1}\right) \left(t\right)$

plug in $4700$ $m$ for ${d}_{x}$ ($4.7$ $k m$ is $4 , 700$ meters)

Then plug in $6.6 m$/$s$ 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 $34.6$ $m$/$s$

Oct 12, 2017

${V}_{\text{final" = 34.6" m/s}}$

#### Explanation:

To find the acceleration, a, use the equation:

$d = {V}_{\text{initial}} t + \frac{1}{2} a {t}^{2}$

where ${V}_{\text{initial" = 6.6" m/s}}$, $d = 4.7 \text{ km" = 4700" m}$, and $t = 3.8 \text{ min" = 228" s}$

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"))

$a = 2 {\left(4700 \text{ m"-(6.6" m/s")(228" s"))/(228" s}\right)}^{2}$

$a = 0.12293 {\text{ m/s}}^{2}$

To find the final speed, ${V}_{\text{final}}$, 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"))#

${V}_{\text{final" = 34.6" m/s}}$