Question

Part AA person who is properly restrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 30 "g's" (1.0g=9.8m/s2).Assuming uniform deceleration at 30 g's, calculate the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 70 km/h .Express your answer to two significant figures and include the appropriate units.

Part B Determine the stopping distances for an automobile going a constant initial speed of 96 km/h and human reaction time of 0.40 s , for an acceleration a=−2.9m/s2. Express your answer using two significant figures.

Answer 1

Part A:

V= 70 km/hr= (70*1000)/(60*60) = 19.44 m/s
a= since 1g=9.8m/s^2
and 30 g=30X9.8 = 294 m/s^2

S=?,U=0m/s
V^2=U^2+2as
19.44^2=0^2+2*294*S

=> S = (19.44^2)/(2*294) = 0.64 m

Part B:

96 km/hr * 1 hr/3600s * 1000 m/km = 26.67 m/s

In 0.40 s, the car traveled 10.67 m.

v² = u² + 2ax

where:
v = final vel = 0
u = initial vel = 26.67 m/s
a = acceleration
x = braking distance

x = (v² - u²)/2a
x = (0 - 26.67²)/2(-2.9)
= 122.64 m braking distance
= 122.64 + 10.67 = 133.31m = total distance