In: Physics
A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components ax(t)=αt2 and ay(t)=β−γt, where α = 2.50 m/s4, β = 9.00 m/s2, and γ = 1.40 m/s3. At t=0 the rocket is at the origin and has velocity v⃗ 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, β, γ, and α. Write the vector v⃗ (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, β, γ, and α. 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"
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