Please help..x and y components..Im stuck!?

Question

On a spacecraft, two engines are turned on for 684 seconds at a moment when the velocity of the craft has x and y components of vox=4370 m/s and voy=6280 m/s. While the engines are firing, the craft undergoes a displacement that has components of x=4.11x10^6 m and y=6.07x10^6 m. Find the x and y components of the craft's acceleration



ok so here are the knowns:


X direction data:
x=4.11x10^6 m
ax= need to know
vx=dont know
vox=4370 m/s
t=684 s

y direction data:
y=6.07x10^6m
ay=need to know
vy=dont know
voy=6280
t=684 s


how would i solve for acceleration? am i using one of the equations of kinematics for constant acceleration in two-dimensional motion or is this not considered constant acceleration? im really confused on where to go from here...

Answer 1

Technically in real life a rocket's acceleration is not constant, but we're going to ignore that little fact for now.

The equation you should be using is ∆x = (1/2)a[x]*t^2 + v[0x]*t
This applies to the y-direction with ∆y = (1/2)a[y]*t^2 + v[0y]*t

∆x is x-displacement and ∆y is y-displacement. a[x] is x-acceleration and a[y] is y-acceleration. v[0x] is x initial velocity and v[0y] is y initial velocity. Now just solve for a[x] and a[y]:

a[x] = 2(∆x / t^2) - 2(v[0x] / t)
a[y] = 2(∆y / t^2) - 2(v[0y] / t)

Soooooo...
a[x] = 4.792 m/s^2
a[y] = 7.586 m/s^2

Answer 2

Leave the y components and find the acceleration along x direction alone.

S = ut + 0.5 a t^2.

4.11 x10^6 = 4370 x 684 + 0.5 a (684) ^2

a along x direction = 4.79m/s^2.

Along Y direction

S = ut + 0.5 a t^2.

6.07 x10^6 = 6280 x 684 +0.5 a (684) ^2

a along y direction = 7.59

Answer 3

So
t= 684 sec

X=V0x t + 0.5 ax t^2
Y=V0y t + 0.5 ay t^2

ax=2 (X - V0x t ) / t^2
ax= 2 ( 4.11 E +6 - 4370 x 684 )/(684 )^2
ax=4.8 m/s^2

ay=2 (Y- V0y t ) / t^2
ay= 7.6 m/s^2

a= sqrt( ax ^2 + ay ^2)
finally a= 9.0 m/s^2