Answer and Explanation:
Given that
v_f = 5 km/s = 5,000 m/s
d = 4 m
v_i = 0 m/s
The computation is shown below:
1. The acceleration in m/s is
Here we use the motio third equation which is
[tex]v_f^2 = v_i^2 + 2ad[/tex]
5000^2 = 0^2 + 2 (a) (4)
So
[tex]a = 3.125 \times 10^{6} m/s^2[/tex]
2. Now acceleration in g is
[tex]= \frac{3.125 \times 10^{6} m/s^2}{9.81}[/tex]
[tex]= 3.18 \times 10^{5}g[/tex]
3. The long of acceleration last is
[tex]t = \frac{v-u}{a}[/tex]
[tex]= \frac{5000 - 0}{3.125 \times 10^{6}}[/tex]
[tex]= 1.6 \times 10^{-3}s[/tex]
4.As we can see that
[tex]3.18 \times 10^{5}[/tex] is smaller than the [tex]4.5 \times 10^{5}g[/tex]
So, it should not be ruled out