Question #3d61e

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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"#