91Ó°ÊÓ

The rate with which a body changes its displacement in unit time is known as velocity. The velocity of a body is a vector quantity, and the unit of velocity is the same as that of speed.

Given data.

The initial speed of the car and the driver is $v_o=95\frac{\mathrm{km}}{\mathrm{h}}$.

The final speed of the car and the driver is $v= 0$as they come to rest.

The driver travels a distance of $\left(x - x_o\right) = 0.80 \mathrm{m}$before coming to rest.

Thus,

Assumption.

Let the acceleration of the driver during the collision be a.

Now, you know that

role="math" localid="1642836990958" $v^2=v_o^2+2a\left(x-x_o\right)…\left(i\right)$.

Now, substituting all the values in equation (i),

$\begin{array}{c}{v}^2 = v_o^2 + 2a \left(x - x_o\right) \\ 0^2 = {\left(26.39 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 + \left[2 × a  × 0.80 \mathrm{m}\right]\\ 1.60\mathrm{m} × a = - {\left(26.39 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2\\ a = -435.3 \frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, the magnitude of the acceleration is $435.3\frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Then,

$\begin{array}{c}a=435.3\frac{\mathrm{m}}{{\mathrm{s}}^2}\\ =435.3 \frac{\mathrm{m}}{{\mathrm{s}}^2} × \frac{1 \mathrm{g}}{9.80 \frac{\mathrm{m}}{{\mathrm{s}}^2}}\\ =44.4 \mathrm{g}\end{array}$

Therefore, the acceleration of the driver was 44.4g during the collision.