Increasing Safety in a Collision Safety experts say that an automobile accident
ID: 2034687 • Letter: I
Question
Increasing Safety in a Collision Safety experts say that an automobile accident is really a succession of three separate collisions. These can be described as follows: (1) the automobile collides with an obstacle and comes to rest, (2) people within the car continue to move forward until they collide with the interior of the car, or are brought to rest by a restraint system like a seatbelt or an air bag, (3) organs within the occupants' bodies continue to move forward until they collide with the body wall and are brought to rest. Not much can be done about the third collision, but the effects of the first two can be mitigated by increasing the distance over which the car and its occupants are brought to rest. For example, the severity of the first collision is reduced by building collapsible "crumple zones" into the body of a car, and by placing compressible collision barriers near dangerous obstacles like bridge supports. The second collision is addressed primarily through the use of seatbelts and air bags. These devices reduce the force that acts on an occupant to survivable levels by increasing the distance over which he or she comes to rest. This is illustrated in the figure(Figure 1), where we see the force exerted on a 65.0-kg driver who slows from an initial speed of 18.0 m/s (lower curve) or 36.0 m/s (upper curve) to rest in a distance ranging from 5.00 cm to 1.00 m Part A A driver who does not wear a seatbelt continues to move forward with a speed of 18.0 m/s (due to inertia) until something solid like the steering wheel is encountered. The driver now comes to rest in a much shorter distance-perhaps only a few centimeters. Find the magnitude of the net force acting on a 65.0 kg driver who is decelerated from 18.0 m/s to rest in 5.00 cm O F 3240 N F= 1.173 × 104 N O F- 2.113 x 105 N OP-4.213 x 105 N Submit Request Answer Provide Feedback Next> Figure 1 of 1> 200,000 Z 150,000 100,000 36.0 m/s 50,000 18.0 m/s 0.200 0.400 0.600 0.800 1.00 Stopping distance (m)Explanation / Answer
Given that,
vi = 18 m/s
vf = 0
m = 65 kg
d = 5 cm = 0.05 m
From the kienematic equation,
vf^2 - vi^2 = 2ad
0 - (18)^2 = 2*a*0.05
a = -3240 m/s^2
Magnitude of net force acting on driver,
F = ma
F = 65*3240
F = 2.11*10^5 N