(a) [tex]-1.5\cdot 10^6 N[/tex]
First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:
[tex]v^2 -u ^2 = 2ad[/tex]
where
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 1.00 cm = 0.01 m is the displacement of the person
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2[/tex]
And the average force on the person is given by
[tex]F=ma[/tex]
with m = 75.0 kg being the mass of the person. Substituting,
[tex]F=(75)(-20000)=-1.5\cdot 10^6 N[/tex]
where the negative sign means the force is opposite to the direction of motion of the person.
b) [tex]-1.0\cdot 10^5 N[/tex]
In this case,
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 15.00 cm = 0.15 m is the displacement of the person with the air bag
So the acceleration is
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2[/tex]
So the average force on the person is
[tex]F=ma=(75)(-1333)=-1.0\cdot 10^5 N[/tex]
The average force on the person if he is stopped by a padded dashboard is -1.5 x 10⁶ N.
The average force on the person if he is stopped by an air bag is -10,000 N.
The given parameters;
The deceleration of the car when the padded dashboard compresses an average of 1.00 cm.
s = 0.01 m
[tex]v^2 = u^2 + 2as\\\\0 = 20^2 + (2\times 0.01)a\\\\0 = 400 + 0.02a\\\\0.02a = -400\\\\a = \frac{-400}{0.02} \\\\a = -20,000 \ m/s^2[/tex]
The average force is calculated as;
F = ma
F = 75 x (-20,000)
F = -1.5 x 10⁶ N.
The deceleration of the car when the air bag compresses an average of 15 cm.
s = 0.15 m
[tex]v^2 = u^2 + 2as\\\\0 = u^2 + 2as\\\\-2as = u^2\\\\a = \frac{u^2}{-2s} = \frac{20^2}{-2\times 0.15} = -1,333.33 \ m/s^2[/tex]
The average force is calculated as follows;
F = ma
F = 75 x (-1,333.33)
F = -10,000 N.
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