A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed about 30 "'s".

Question

Assuming uniform deceleration of this value, 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 85

Answer 1

vf^2=vi^2+2ad

for a,30*9.8 m/s^2; vf=0, vi= convert to m/s
solve for d

Answer 2

A ball is tossed vertically upward from the ground and caught at a height of 15m. After 1.85seconds.
a)what is the velocity with which it was throuwn?
b)What maximum height does it reach.
c)whatt was the velocity when it was caught?

Answer 3

A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed about 30 "'s".
Assuming uniform deceleration of this value, 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 85.
can you please help me am stuck

Answer 4

Since we want our units to be in meters and seconds, lets first convert 85 km/hr to m/s.

85 km 1000 m 1 hr
---------- x ------------- x -------------- = 23.61 m/s
1 hr 1 km 3600 sec

Next, lets convert the 30 "g's" to m/s^2, knowing that 1.0 g = 9.8 m/s^2

30 g 9.8 m/s^2
-------- x --------------- = 294 m/s^2
1 1 g

Now lets fill in the variables we know:

Initial velocity (Vo) = 23.61 m/s
End velocity (V) = 0 m/s (since the car has stopped at this point)
Initial position (Xo) = 0 m
End position (X) = ?
Acceleration (a) = -294 m/s^2 (The # is negative since we are decelerating)

Now all we have to do is plug our variables into an equation to find our end position. For this we will use the following equation and complete the following steps:

V^2 = Vo^2 + 2a(x-xo)
0 = (23.61)^2 + 2(-294) (X-0)
0 = 557.48 - 588x
-557.48 = -588x
x = 0.95 m

So our answer is that the car needs to be designed to go 0.95 m. Hopefully this helps!