Respuesta :
We use the Newtons law,
[tex]F=m a[/tex] (A)
Equation of motion,
[tex]v = u + at[/tex] (B)
Here, v is final velocity , u is initial velocity and a is acceleration.
Given [tex]u = 110 \ km/h = 110 \frac{10^3 m}{60 \times 60 \ s} =30.55 \ m/s, t = 0.14 s[/tex] .
As car coming to stop, so final velocity becomes zero i.e [tex]v = 0[/tex].
Substituting these values in equation (B), we get
[tex]0 = 33 . 55 \ m/s + a (0.14 \ s ) \\\\ a = \frac{ (-33.50 m/s )}{0.14 s } =- 239 .64 \ m/s^2[/tex]
Here, negative sign means, deceleration to rest.
As given the mass of passenger, m = 60 kg.
Therefore, from equation (A),
[tex]F = (60 kg) (-239 .64 \ m/s^2) = -14378.57 N[/tex]
Thus , the force seat belt exert on the passenger during the collision is [tex]-14378.57 \ N[/tex]
