A car traveling at 95 km/h strikes a tree. The front end of the car compresses, and the driver comes to rest after traveling 0.80 m. What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of ‘g’s’, where 1.00 g = 9.80 m/s².

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" $v^2=v_o^2+2a\left(x-x_o\right)\dots \left(i\right)$.

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

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

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

Then,

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

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