How Does a Faulty Model Rocket's Motion Change Over Time?

[mohd22]

1. A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components a_{x}(t)= \alpha t^{2} and a_{y}(t)= \beta - \gamma t, where \alpha = 2.50 {\rm m}/{\rm s}^{4}, \beta = 9.00 {\rm m}/{\rm s}^{2}, and \gamma = 1.40 {\rm m}/{\rm s}^{3} . At t = 0 the rocket is at the origin and has velocity {\vec{v}}_{0} = {v}_{0x} \hat{ i } + v_{0y} \hat{ j } with v_{0x} = 1.00 {\rm m}/{\rm s} and v_{0y} = 7.00 {\rm m}/{\rm s} .



2. Calculate the velocity vector as a function of time.
Express your answer in terms of v_0x, v_0y, beta, gamma, and alpha. Write the vector \vec{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 v_0x, v_0y, beta, gamma, and alpha. Write the vector r(t)_vec in the form r(t)_x, r(t)_y where the x and y components are separated by a comma

Your answer should be an expression, not an equation.

 

[CompuChip]

Since your post contains LaTeX code, you might find the [ tex] and [ itex] tags useful (without the space), like so:
a_{x}(t)= \alpha t^{2} (display mode for use on separate line)
a_{x}(t)= \alpha t^{2} (text mode for inline use)
- you can click on them to see the code.

As for your question, you are given the acceleration and asked the velocity. That seems rather straightforward to me. What did you try already?