Answer:
19129.5 N
1.1831 m
Explanation:
m = Mass of person = 65 kg
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration = 30g
g = Acceleration due to gravity = 9.81 m/s²
Force is given by
[tex]F=ma\\\Rightarrow F=65\times 30\times 9.81\\\Rightarrow F=19129.5\ N[/tex]
The force on the person is 19129.5 N
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-\left(95\times \dfrac{1000}{3600}\right)^2}{2\times -9.81\times 30}\\\Rightarrow s=1.1831\ m[/tex]
The distance traveled if brought to rest is 1.1831 m
The distance the person moves before coming to rest depends on the
deceleration.
(a) The force on the person is 19,129.5 N
(b) The distance traveled before coming to rest is approximately 1.183 m
Reasons:
(a) The force on the person = Mass × Acceleration
The deceleration of the person = 30 g's
The person's mass, m = 65 - kg
Therefore;
The force on the person = 65 kg × 30 × 9.81 m/s² = 19,129.5 N
(b) Initial speed, u = 95 km/h = 26.38889 m/s
Deceleration, a = 30 g's
We get;
v² = u² - 2·a·s
When v = 0, we get;
0² = 26.38889² - 2 × 30 × 9.81 × s
2 × 30 × 9.81 × s = 26.38889²
[tex]s = \dfrac{26.3889^2}{2 \times 30 \times 9.81} = 1.18310325[/tex]
The distance traveled before coming to rest, s ≈ 1.183 m
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