## G-Force

On 13 July 1977, British racing driver David Purley survived one of the highest G-load crashes in history during a Formula 1 race. In about a span of 0.66 m, his car went from travelling 173 km/h to zero. How many g loads did his body endure?

Expand Hint
$$v^2=v_0^2+2a_0(s-s_0)$$$where $$v$$ is the velocity along the direction of travel, $$v_0$$ is the velocity at time $$t_0$$ , $$a_0$$ is constant acceleration, $$s$$ is the displacement at time $$t$$ along the line of travel, and $$s_0$$ is the displacement at time $$t_0$$ . Hint 2 $$173\frac{km}{hr}\cdot \frac{1hr}{60min}\cdot\frac{1min}{60s}\cdot\frac{1,000m}{1km}=48.056\:\frac{m}{s}$$$
For constant acceleration, the equation for velocity as a function of position:
$$v^2=v_0^2+2a_0(s-s_0)$$$where $$v$$ is the velocity along the direction of travel, $$v_0$$ is the velocity at time $$t_0$$ , $$a_0$$ is constant acceleration, $$s$$ is the displacement at time $$t$$ along the line of travel, and $$s_0$$ is the displacement at time $$t_0$$ . First, let’s convert the initial speed into $$m/s$$ : $$173\frac{km}{hr}\cdot \frac{1hr}{60min}\cdot\frac{1min}{60s}\cdot\frac{1,000m}{1km}=48.056\:\frac{m}{s}$$$
Solving for deacceleration:
$$a_0=\frac{(v^2-v_0^2)}{2(s-s_0)}=\frac{0-(48.056\frac{m}{s})^2}{2(0.66m)}=\frac{-2,309.34\frac{m}{s^2}}{1.32}=-1,749.5\:\frac{m}{s^2}$$$Thus, the amount of g loads David Purley endured: $$\frac{1,749.5m/s^2}{9.8m/s^2}=179\:g$$$
For comparison, astronauts typically only experience 3 g when leaving the launch pad.
179 g