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 = ( 4 1 t 2 ) , m/s 2 , where t is in seconds, determine the magnitude of the rockets velocity and acceleration when t = 10 s.

The velocity is 1003.5 m/s  and the acceleration is 103.1 m/.

Explanation:

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 =

dv = adt

∫dv = ∫adt

 =  

v = .

Velocity is:

v =

dx=vdt

Again, integrate both sides:

∫dx = ∫vdt

=

x =

Substitute our x equation into our parameter equation of y.

y = [120()x] m

y = [120()()() ]

(take the square root of both sides and simplify)

y = 50

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 = 50

vy= y = 100t

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

vx =

vy = 100t m/s

At t = 10 s:

vx = = 83.3 m/s

vy  = 100(10)=1000 m/s

The magnitude of velocity is:

v =

v =

To figure out the acceleration, we need to figure out ay which can be found by taking the derivative of the vy equation,

vy = 100t m/s

ay = vy = 100 m/

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

ax =

ay = 100 m/

At t = 10 s:

ax =

ay = 100 m/

The magnitude of acceleration is equal to:

a =

a =

a= 103.1 m/

Final Answers:

v = 1003.5 m/s

a= 103.1 m/

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