## 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**

**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**

**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, letâ€™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

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