A car traveling at 90 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.72 m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g = 9.80 m/s2
Vi= 90km/h = 25m/sec
Vf = 0 m/sec
S= 0.72 m
a= ?
SOL: By 3rd equation of motion, we have
2aS = Vf2 - Vi2
a = -434 m/sec2
To find the average acceleration of the driver during the collision, we first need to calculate the time it took for the driver to come to rest.
We can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s because the driver comes to rest)
u = initial velocity (90 km/h = 25 m/s)
s = displacement (0.72 m)
a = acceleration (unknown)
Rearranging the equation to solve for acceleration:
a = (v^2 - u^2) / (2s)
Substituting the known values:
a = (0 - (25)^2) / (2 * 0.72)
Simplifying:
a = -625 / 1.44
a = -434.03 m/s^2
Now, to express the acceleration in terms of "g's," we divide the obtained acceleration by the acceleration due to gravity (9.80 m/s^2):
acceleration in "g's" = -434.03 / 9.80
acceleration in "g's" ≈ -44.31 g's
Therefore, the average acceleration of the driver during the collision is approximately -44.31 g's. Note that the negative sign indicates that the acceleration is in the opposite direction of the car's initial motion.