A faulty model rocket moves in the xy -plane (the positive y -direction is verti

Question

A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components ax(t)=t2 and ay(t)=t, where = 2.50 m/s4, = 9.00 m/s2, and = 1.40 m/s3. At t=0 the rocket is at the origin and has velocity v 0=v0xi^+v0yj^ with v0x = 1.00 m/sand v0y = 7.00 m/s.

Calculate the velocity vector as a function of time.Express your answer in terms of v0x, v0y, , , and . Write the vector v (t) in the form v(t)x, v(t)y, where the x and y components are separated by a comma.

Calculate the position vector as a function of time.Express your answer in terms of v0x, v0y, , , and . Write the vector r(t) in the form r(t)x, r(t)ywhere the x and y components are separated by a comma.

What is the maximum height reached by the rocket?

What is the horizontal displacement of the rocket when it returns to y=0?

Explanation / Answer

V(t)x= Vox+ax(t)*t = Vox+at2*t = Vox+at3

V(t)y= Voy+ay(t)*t = Voy+(-Yt)*t = Voy+ t-Yt2

V(t) = V(t)x+V(t)y = [ (Vox+at3) i^ , (Voy+t-Yt2) j^ ]

r(t)x= [V(t)x]dt = (Vox+at3) dt = (Vox*t+at4/4)

r(t)y= [V(t)y]dt = (Vox+t-Yt2) dt = (Voy*t+t2/2 –Yt3/3)

r(t) = r(t)x+r(t)y = [(Vox*t+at4/4)i^ , (Voy*t+t2/2 –Yt3/3)j^]

Use kinematic equation

V(t)y = Voy+ay(t)*t

V(t)y = Voy + Voy+ t-Yt2

at Hmax V(t)y = 0m/s

Plug values,

0 = 7 + 9t-1.4t2          => t= 7.13s

r(t)y at t= 7.13s

r(t)y = (Voy*t+t2/2 –Yt3/3) = (7*7.13+ (9*7.132/2) – ((1.4*7.133)/3) ) = 109.5 m

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