A rocket is fired from rest at x = 0

A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by y^2=[120(10^3)x] m. If the x component of acceleration is a_x=(\dfrac{1}{4}t^2) m/s^2, where t is in seconds, determine the magnitude of the rocket’s velocity and acceleration when t = 10 s.


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From the question, we are given the parameter equation of y. We need to find the parameter equation of x. To find it, we will need to integrate the x-component acceleration equation given to us, twice. Acceleration is:




Integrate both sides:

\,\displaystyle \int dv=\int a\,dt

\,\displaystyle \int^{v_x}_{0}dv=\int^{t}_{0}\dfrac{1}{4}t^2\,dt

v_x=\dfrac{1}{12}t^3 m/s


Velocity is:




Again, integrate both sides:

\,\displaystyle \int dx=\int v\,dt

\,\displaystyle \int^{x}_{0}dx=\int^{t}_{0}\dfrac{1}{12}t^3\,dt

x=\dfrac{1}{48}t^4 m


Substitute our x equation into our parameter equation of y.

y^2=[120(10^3)x] m


(take the square root of both sides and simplify)



Now that we can represent our equation with respect to time, we can take the derivative to figure out the velocity. Remember that taking the derivative of a position function gives us the velocity function.




Let us write down the two equations for velocity we found:

v_x=\dfrac{1}{12}t^3 m/s

v_y=100t m/s


At t = 10 s:

v_x=\dfrac{1}{12}(10)^3=83.3 m/s

v_y=100(10)=1000 m/s


The magnitude of velocity is:


v=\sqrt{(83.3)^2+(1000)^2}=1003.5 m/s


To figure out the acceleration, we need to figure out a_y which can be found by taking the derivative of the v_y equation.

v_y=100t m/s

a_y=\dot{v_y}=100 m/s^2


Since a_x is given to us in the question, we have the following:

a_x=(\dfrac{1}{4}t^2) m/s^2

a_y=100 m/s^2


At t = 10 s:

a_x=(\dfrac{1}{4}(10)^2)=25 m/s^2

a_y=100 m/s^2


The magnitude of acceleration is equal to:


a=\sqrt{(25)^2+{100}^2}=103.1 m/s^2


Final Answers:

v=1003.5 m/s

a=103.1 m/s^2

This question can be found in Engineering Mechanics: Dynamics (SI edition), 13th edition, chapter 12, question 12-82.

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