# 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. Designed by Freepik

#### Solution:

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:

$a=\dfrac{dv}{dt}$

$dv=a\,dt$

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:

$v=\dfrac{dx}{dt}$

$dx=v\,dt$

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

$y^2=[120(10^3)(\dfrac{1}{48})(t^4)]$

(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.

$y=50t^2$

$v_y=\dot{y}=100t$

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:

$v=\sqrt{(v_x)^2+(v_y)^2}$

$v=\sqrt{(83.3)^2+(1000)^2}=1003.5$ 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:

$a=\sqrt{(a_x)^2+(a_y)^2}$

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

$v=1003.5$ m/s
$a=103.1$ m/$s^2$