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.
Solution:
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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:
dv=a\,dt
Integrate both sides:
\,\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:
dx=v\,dt
Again, integrate both sides:
\,\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)(\dfrac{1}{48})(t^4)]
(take the square root of both sides and simplify)
y=50t^2
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.
v_y=\dot{y}=100t
Let us write down the two equations for velocity we found:
v_y=100t m/s
At t = 10 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.
a_y=\dot{v_y}=100 m/s^2
Since a_x is given to us in the question, we have the following:
a_y=100 m/s^2
At t = 10 s:
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:
a=103.1 m/s^2