Question #3d61e

Question

1732 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-12T01:37:25

$34.6$ $m$ / $s$

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 · 2017-10-12T01:45:56

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

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:

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

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

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