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A particle position as a function of time t is given by r=(5.0t+6.0t^2)mi+(7.0-3.0t^3)mj. At t=5s find the magnitude and direction of the particles displecment vector delta(r) relative to the point r0=(0.0i+7.0j)m

Respuesta :

AL2006
Horizontal component:

x(t) = 5t + 6t²

x(5) = 25 + 6(25) = 175 m 

Vertical component:

y(t) = 7 - 3t³

y(5) = 7 - 3(125) = -368 m

Components combined:

r(5)  =  (175 m) i  -  (368 m) j

Taken relative to the point  r(0) = (0 i  +  7 j) m ,
the displacement vector when t=5 is

           (175 m) i  -  (375 m) j .

Its magnitude is  √(175² + 375²) 

                     =  √(30,625 + 140,625)

                     =   √171,250  =  413.823...  m      (rounded)

Its direction is   tan⁻¹(-375/175)  =  tan⁻¹(-2.14285...)  =  - 64.98°       

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