Question

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.

Answer 1

Given data,

The initial speed of the car and the driver is v₀ = 95 km/h.

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

The driver travels a distance of (x  -  x₀)  =  0.80 m before coming to rest.

Thus,

Assumption.

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

Now, you know that

\(v^2 = v_0^2 + 2a(x - x _0) \quad ....(i)\)

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

\(v^2 = v_0^2 + 2a(x - x _0) \)

\(0^2 = (26.39\ m/s)^2 + [2a \times a \times 0.80 m]\)

\(1.60 m \times a = -(26.39 \ m/s)^2\)

\(a = -435 .3 \ m/{s^2}\)

Therefore, the magnitude of the acceleration is 435.3 m/s².

Then,

\(a = 435.3 \ m/s^2\)

\(= 435.3 \ m/s^2 \times \frac {1g}{9.80\ m/s^2}\)

\(= 44.4 g\)

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