Respuesta :
As we know by the kinematics
[tex]\delta x = v_x * t + \frac{1}{2} at^2[/tex]
[tex]4.11106 = 4370* 684 + \frac{1}{2}a_x*684^2[/tex]
[tex]4.11106 = 2989080 + 233928*a_x[/tex]
[tex]a_x = -12.78 m/s^2[/tex]
Similarly for Y direction
[tex]\delta y = v_y * t + \frac{1}{2} at^2[/tex]
[tex]6.07106 = 6280* 684 + \frac{1}{2}a_y*684^2[/tex]
[tex]6.07106 = 4295520 + 233928*a_y[/tex]
[tex]a_y = -18.36 m/s^2[/tex]
Answer:
X - Component of acceleration = 4.79 m/s²
Y - Component of acceleration = 7.59 m/s²
Explanation:
We have equation of motion, s = ut + 0.5 at²
X component:-
Initial velocity, u = 4370 m/s
Time, t = 684 s
Displacement, s = 4.11 x 10⁶ m
Substituting,
s = ut + 0.5 at²
4.11 x 10⁶ = 4370 x 684 + 0.5 x a x 684²
a = 4.79 m/s²
X - Component of acceleration = 4.79 m/s²
Y component:-
Initial velocity, u = 6280 m/s
Time, t = 684 s
Displacement, s = 6.07 x 10⁶ m
Substituting,
s = ut + 0.5 at²
6.07 x 10⁶ = 6280 x 684 + 0.5 x a x 684²
a = 7.59 m/s²
Y - Component of acceleration = 7.59 m/s²