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