A car traveling at 120 strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.90 .
What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of "'s," where 1.00 = 9.80 .
Lets get acceleration.
Vf^2 = Vi^2 + 2a*d
Vf = Final velocity
Vi = Initial veloctiy
a = acceleration
d = distance traveled
0 = (120)^2 + 2a(0.90)
a = -8000
Ava = (Vf - Vi/(Tf - Ti)
First must find time.
Ava = average acceleration
Vf = Vi + a*T
Tf = final time, Ti = initial time
0 = 120 -8000*T
T = 0.015
Ava = (0 - 120)/(0 - 0.015)
Ava = 8000
8000 * 9.80
Magnitude of Ava = 78,400 or 8000s
To find the magnitude of the average acceleration of the driver during the collision, we can use the formula:
Acceleration = (Final Velocity - Initial Velocity) / Time
Given:
Initial Velocity (u) = 120 m/s (the speed of the car)
Final Velocity (v) = 0 m/s (the driver comes to rest)
Time (t) = 0.90 s
Substituting the values into the formula:
Acceleration = (0 - 120) / 0.90
Now we can calculate the average acceleration:
Acceleration = -120 / 0.90
Acceleration = -133.33 m/s^2
Since the question asks for the answer to be expressed in terms of g's, we need to convert the acceleration to g-units. 1 g = 9.80 m/s^2.
Acceleration in g's = (-133.33 m/s^2) / (9.80 m/s^2)
Acceleration in g's = -13.61 g's
Therefore, the magnitude of the average acceleration of the driver during the collision is 13.61 g's.