Answer
x-component of the acceleration $=4.79\space m/s^{2}$
y-component of the acceleration $=7.59\space m/s^{2}$
Work Step by Step
Here we use equation 3.5a,b $S=ut+\frac{1}{2}at^{2}$ in horizontal, and vertical directions to find the x, and y components of its acceleration.
$\rightarrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$4.11\times 10^{6}m=4370\times684\space m+\frac{1}{2}a_{x}(684\space s)^{2}$
$a_{x}=\frac{2(4.11\times 10^{6}m-4370\times684\space m)}{(684\space s)^{2}}=4.79\space m/s^{2}$
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$6.07\times 10^{6}m=6280\times684\space m+\frac{1}{2}a_{y}(684\space s)^{2}$
$a_{y}=\frac{2(6.07\times 10^{6}m-6280\times684\space m)}{(684\space s)^{2}}=7.59\space m/s^{2}$
