The position of a 2.75×10^5 N training helicopter under test is given by r =(0.020m/s 3 )t^3 i ^ +(2.2m/s)tj ^ −(0.060m/s 2 )t^2 k ^. find the net force of the helicopter at t=5.0s

Answer 1

r =(0.020m/s³)t ³ i ^ +(2.2m/s)tj ^ −(0.060m/s² )t²k ^.
v= dr/dt =(3•0.020m/s³ ) •t²• i ^ +(2.2m/s) • j ^ −2• (0.060m/s² )t•k ^.
a=d²r/dt²=(6•0.020m/s ³) •t i ^ -2• (0.060m/s² )•k ^.
F=ma =2.75•10⁵•{(6•0.020m/s ³) •t i ^ -2• (0.060m/s² )•k ^.}
t=5 s
F₅=2.75•10⁵•{(6•0.020m/s ³) •5• i ^ -2• (0.060m/s² )•k ^.}=
=165000 i ^ - 33000•k ^.

Answer 2

the actual answer for two sig figs are 1.7x10^4*i^-3.3x10^3*k^1