Thus, the distance traveled by automobile is [tex]=1.31m[/tex]
The mass of the person is [tex]m=65kg[/tex]
The deceleration of the automobile is [tex]a=30g[/tex]
The acceleration due to gravity is [tex]g=9.81m/s.[/tex]
The initial velocity of the automobile is [tex]v_{i}=100 km/h*\frac{1m/s}{3.6km/h} =27.78m/s[/tex]
Determination of the force exerted on the person:
The force applied to the person can be calculated using Newton's second law as follows:
[tex]F=ma[/tex]
[tex]30g[/tex]Here, m denotes the person's mass, while a denotes the car's acceleration. Because the driver is moving at 30g, the car's acceleration is assumed to be positive.
Substitute the values as 70 kg for [tex]m,(30g)[/tex] for a, and[tex]9.81m/s^{2}[/tex]
for g in the above equation.
[tex]F=70kg*30g[/tex]
[tex]=70kg*30g*9.81m/s^{2} (\frac{1N}{1kg*m/s^{2} })[/tex]
[tex]=20,601N[/tex]
Thus, the force acting on the person is [tex]=20,601N[/tex]
Determination of the distance traveled by automobile:
The final velocity of the car can be calculated using Newton's third law and represented as follows:
[tex]v_{f} ^{2}=v_{i} ^{2}+2ad[/tex]
[tex]d=\frac{v_{f} ^{2}=v_{i} ^{2}}{2a}[/tex]
Here [tex]v_{f}[/tex] is the final velocity of the automobile. The final velocity of the automobile becomes 0 m/s as the automobile comes to rest. The acceleration is taken to be negative because the automobile is decelerating at [tex]30g[/tex]
Substitute the values as [tex]0 m/s[/tex] for[tex]v_{f} =27.78m/s[/tex], for[tex]v_{i} ,(-30g)[/tex], for a, and [tex]9.81m/s^{2}[/tex]for g in the above equation.
[tex]d=\frac{(0 m/s)^{2} - (27.78m/s)^{2} }{2*(-30)*9.81m/s^{2} }[/tex]
[tex]=1.31m[/tex]
Thus, the distance traveled by automobile is[tex]=1.31m[/tex]
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