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)=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 v 0=v0xi^+v0yj^ with v0x = 1.00 m/s and 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)y where 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.

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

Answer 1

a) Vy(t) = Vox +4 +3 Vy(t) = Voy + Bt-2ť VA) = vxt) T + y(t)} 7(4) = (vox tat, bytBt- Ft) b) r(t) = Voxt + 2 th, rit)y = voyt + Ft -2 +3 0 No=(xxt +8th, byt + B +_ 843) e set Vy =0 yields a quadratic int, 10 = Voy + Bt-rt, which has as the positive solution. t=% [R+VB" + 2 vej t = tyo [9 +594 264) 140) = 13:59 5 use this time in the expression © which gives a maximum height of 341 m ) Y(+) 400 t (m) 300 18104 28104 3x107 uxloy a(t) (m) e) y zo gives o = voyt + f + 8 +3 Šť- Bt-voy = 0. By solving Quadratic equatim. The positive solution is t = 20.735 For this x = 3-85X10" m