Solution Found!
Answer: CALC A faulty model rocket moves in the xy-plane
Chapter 3, Problem 42P(choose chapter or problem)
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 \(\alpha=2.50\mathrm{\ m}/\mathrm{s}^4,\ \beta=9.00\mathrm{\ m}/\mathrm{s}^2\), and \(\gamma=1.40\mathrm{\ m}/\mathrm{s}^3\). At t = 0 the rocket is at the origin and has velocity \(\vec{\boldsymbol{v}}_{0}=v_{0 x} \hat{\imath}+v_{0 y} \hat{\jmath}\) with \(v_{0x}=1.00\mathrm{\ m}/\mathrm{s}\) and \(v_{0y}=7.00\mathrm{\ m}/\mathrm{s}\). (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) Sketch the path of the rocket. (d) What is the horizontal displacement of the rocket when it returns to y = 0?
Questions & Answers
QUESTION:
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 \(\alpha=2.50\mathrm{\ m}/\mathrm{s}^4,\ \beta=9.00\mathrm{\ m}/\mathrm{s}^2\), and \(\gamma=1.40\mathrm{\ m}/\mathrm{s}^3\). At t = 0 the rocket is at the origin and has velocity \(\vec{\boldsymbol{v}}_{0}=v_{0 x} \hat{\imath}+v_{0 y} \hat{\jmath}\) with \(v_{0x}=1.00\mathrm{\ m}/\mathrm{s}\) and \(v_{0y}=7.00\mathrm{\ m}/\mathrm{s}\). (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) Sketch the path of the rocket. (d) What is the horizontal displacement of the rocket when it returns to y = 0?
ANSWER:Solution 42P
The x component of the acceleration is,
The y component of the acceleration is,
Where,
.
The x component of the velocity is,
The y component of the velocity is,
a)
So, according to the equation of motion the velocity as a function of time would be,