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.

A)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.

B)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.

C)What is the maximum height reached by the rocket?

D)Sketch the path of the rocket( x-axies presents x,m ;10000,20000,30000 , 40000)(y-axies presents y,m ; 100,200,300,400)<--- plz make the the point clear so I sketch it right

E)What is the horizontal displacement of the rocket when it returns to y=0?(in KM)

"clear answers plz"

Explanation / Answer

ax(t) = [alpha]*t^2

= (2.5m/s^4) * t^2

vx(t) = .8333m/s^4 * t^3 + C //integrate int(ax(t),dt) = vx(t) + C

1m/s = .8333m/s^4 * t^3 + C

C = 1m/s

vx(t) = (.8333m/s^4) * t^3 + 1m/s

sx(t) = (.20833m/s^4) * t^4 + (1m/s) * t + C //C=0 b/c s(0)=0

sx(t) = (.20833m/s^4) * t^4 + (1m/s) * t ****************************************

ay(t) = [beta] – [gamma]*t

= 9m/s^2 – (1.4m/s^3) * t

vx(t) = (9m/s^2) * t – (.7m/s^3) * t^2 + C //integrate int(ay(t),dt) = vy(t) + C

7m/s = (9m/s^2) * t – (.7m/s^3) * t^2 + C

C = 7m/s

vy(t) = (9m/s^2) * t – (.7m/s^3) * t^2 + 7m/s

sy(t) = (4.5m/s^2) * t^2 – (.23333m/s^3) * t^3 + (7m/s) * t ***************************************

Vvect(t) = (.8333m/s^4 * t^3 + 1m/s)*i + ((9m/s^2) * t – (.7m/s^3) * t^2 + 7m/s)*j

Svect(t) = ((.20833m/s^4) * t^4 + (1m/s) * t)*i + ((4.5m/s^2) * t^2 – (.23333m/s^3) * t^3 + (7m/s) * t)*j

b.) Max height is reached when vy = 0.

vy(t) = (9m/s^2) * t – (.7m/s^3) * t^2 + 7m/s

0m/s = (9m/s^2) * t – (.7m/s^3) * t^2 + 7m/s

t = 13.59s

sy(13.59s) = (4.5m/s^2) * (13.59s)^2 – (.23333m/s^3) * (13.59s)^3 + (7m/s) * (13.59s)

= 831.10m – 585.64m + 95.13m

= 340.59m

d.) sy(t) = (4.5m/s^2) * t^2 – (.23333m/s^3) * t^3 + (7m/s) * t

0 = (4.5m/s^2) * t^2 – (.23333m/s^3) * t^3 + (7m/s) * t

t = 20.733s //used TI-SOLVE

sx(20.733s) = (.20833m/s^4) * (20.733s)^4 + (1m/s) * (20.733s)

= 38494.65m + 20.733m

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