Answer:
[tex]a_x=12.8m/s^{2}[/tex]
[tex]a_y=16.8m/s^{2}[/tex]
Explanation:
To solve this problem, we need to make use of Newton's Second Law of motion which states that the rate of change of momentum of an object is directly proportional to the force applied to the body. Mathematically, this is
[tex]F=\frac{\Delta P}{\Delta t}[/tex],
where [tex]F[/tex] is the force applied to the object, [tex]\Delta P[/tex] is the change in momentum and, [tex]\Delta t[/tex] is the time it takes for the momentum to change.
We can simplify this equation remembering that
[tex]F=ma\\\Delta P= P_{f}-P_{i}=mv_{f}-mv_{i}=m(v_{f}-v_{i})[/tex]
where [tex]m[/tex] is the mass, [tex]a[/tex] is the acceleration, [tex]v[/tex] is the velocity and, [tex]i[/tex] and [tex]f[/tex] stands for initial and final, respectively.
Now, Newton's Second Law becomes
[tex]ma=\frac{m(v_f-v_i)}{\Delta t}\\\\\\a=\frac{(v_f-v_i)}{\Delta t}[/tex].
Listing the data we have:
1. [tex]v_{i}=13km/s=13000m/s[/tex]
2.[tex]v_f=23km/s=23000m/s[/tex] at [tex]26^{\circ}[/tex]
3. [tex]\Delta t=10min=600s[/tex].
Considering x-axis:
[tex]a_x=\frac{(v_f_x-v_i_x)}{\Delta t}[/tex]
At the begining, the motion was along the x-axis, and at the end, the asteroid was moving at
[tex]v_f_x=v_f*cos(\theta)=23000cos(26)=20672.26m/s[/tex],
so
[tex]a_x=\frac{(20672.26-13000)}{600}\\\\a_x=12.8m/s^{2}[/tex]
Now, considering y_axis
[tex]a_y=\frac{(v_f_y-v_i_y)}{\Delta t}[/tex].
At the begining, there was no initial velocity, this means [tex]v_i_y=0[/tex]. At the end, the velocity was [tex]v_f_y=v_fsin(\theta)=26000sin(26)=10082.54m/s[/tex], so
[tex]a_y=\frac{(10082.54)}{600}[/tex]
[tex]a_y=16.8m/s^{2}[/tex]
