A car traveling at 100 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.85 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.
To find the magnitude of the average acceleration of the driver during the collision, we will use the equation of motion:
v² = u² + 2aΔx
where:
v = final velocity (0 m/s, because the driver comes to a rest)
u = initial velocity (100 km/h)
a = average acceleration during the collision
Δx = displacement (0.85 m)
First, we need to convert the initial velocity from km/h to m/s:
100 km/h * (1000 m/1 km) * (1 hr/3600 s) = 27.8 m/s
Substituting the known values into the equation of motion:
0² = (27.8 m/s)² + 2a(0.85 m)
Expanding and simplifying the equation:
0 = 771.24 m²/s² + 1.7a
Rearranging the equation to solve for a:
1.7a = -771.24 m²/s²
a = -771.24 m²/s² / 1.7
Calculating the value of a:
a ≈ -453.14 m²/s²
Finally, let's express the acceleration in terms of "g's":
1.00 g = 9.80 m/s²
So, to find the magnitude of the average acceleration in terms of g's, we divide the acceleration by the acceleration due to gravity:
a = -453.14 m²/s² / (9.8 m/s²/g)
a ≈ -46.23 g's
Therefore, the magnitude of the average acceleration of the driver during the collision is approximately 46.23 g's.
v initial = 100000 m / 3600 s
v final = 0
v ave = (vi + vf) / 2
time = .85 m / (v ave)
accel = vi / time