Answer:
(a) The acceleration oF the car is [tex]-1764.7 m/s^2[/tex] or -180.1 g’s.
(b) The magnitude of the force on the driver's body is 123529 Newtons
Explanation:
Given that:
The initial velocity of the car before the impact, u=172.8 km/h
Or, [tex]u =172.8 \times \frac {1000}{3600} m/s= 48m/s[/tex]
As the car came to a complete stop after the collision, so the final velocity of the car, v=0.
The time required to stop the car completely, [tex]t=2.72\times 10^{-2} s[/tex]
(a) Assuming the acceleration is constant, so
Acceleration, [tex]a= (v-u)/t[/tex]
So, [tex]a=\frac {0-48}{2.72\times 10^{-2}}=-1764.7 m/s^2[/tex]
Now, expressing the acceleration in the unit od "g's",
[tex]a=-1764.7\times \frac{g}{g} \\\\\Rightarrow a=-1764.7 \;m/s^2\times \frac{g}{9.8\; m/s^2} \\\\\Rightarrow a=-180.1 g[/tex]
Hence, the acceleration od the car is -1764.7 [tex]m/s^2[/tex] or -180.1 g’s.
(b) Mass of the driver of the car, m=70.0 kg
The acceleration of the body of the driver is the same as the acceleration of the car.
So, the acceleration faced by the driver, [tex]a= -1764.7 m/s^2[/tex]
As forec = mass x acceleration, so
The force on the driver's body = 70 x (-1764.7)= -123529 Newtons.
The negative sign means the force is in the opposite direction of the direction of motion of the car.
Hence, the magnitude of the force on the driver's body is 123529 Newtons