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

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/s and v0y = 7.00 m/s.

1. Calculate the velocity vector as a function of time.

2. Calculate the position 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.

What is the maximum height reached by the rocket?

What is the horizontal displacement of the rocket when it returns to y=0? 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/s and v0y = 7.00 m/s.

1. Calculate the velocity vector as a function of time.

2. Calculate the position 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.

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

it is just a matter of integration and using initial conditions
since in general dv/dt = a
it implies v = integral a dt

v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get
v(t)_x = alpha t^3 / 3 + v_{0x}

similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y}
v(t)_y = beta t - gamma t^2 / 2 + v_{0y}

so the velocity vector as a function of time
vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]

similarly you should integrate to find position vector since dr/dt = v
r = integral of v dt

r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases

r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume

r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]

recall that a = dv/dt or dv = a dt

we find vy(t) by integrating ay dt

v(t) = Integral(2.5 - 1.4 t dt) = 2.5 t - 0.7t^2 +K where K is a constant of integration

since vy=7 when t=0, K=7 and vy(t)= 2.5t-0.7t^2+7

max height occurs when vy=0; solve the quadratic 0=2.5t-0.7t^2 +7 to find this time

now, integrate vy(t) dt to find y(t)

y(t) = 1.25t^2 - 0.233t^3 + 7t +K" where K" is a constant of integration

since y=0 at t=0, K"=0 and y(t) = 1.25t^2-0.233t^3+7t

evaluate this at the time at which max ht occurs to find the value of max ht

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