Average G-Force Calculation for Abrupt Deceleration from 30mph in 1m Distance

[bond535]

Homework Statement

Determine the average number of g's experienced by someone abruptly coming to rest from an initial velocity of 30mph (such as in a car crash). Take the distance involved in coming to rest to be 1 m.

Homework Equations

V=Vo+at
ΔS=VoT+1/2at^2
V^2=Vo^2+2aΔS

The Attempt at a Solution

t=?
a=?
Vo= 30 mph
V=?
ΔS=?

Thank you, I'm really stuck and any help will be deeply appreciate it.

 

[lewando]

You know V_final = 0 and ΔS (change in distance) = 1m. does that help?

 

[bond535]

so...
t=?
a=?
Vo= 30 mph
V=0
ΔS=1m

V^2=Vo^2+2aΔS
0=(30mph)^2+2a(1m)
which resulted in a= -450mph

I know g= 9.8m/s ... so I'm assuming I have to change the -450mph to m/s by diving -450/3600s = 0.125m/s

I really don't know what to do next or if I'm in the right track...

 

[lewando]

V^2=Vo^2+2aΔS
0=(30mph)^2+2a(1m)
which resulted in a= -450mph

a should be in units of m/s²

(convert mph to m/s)

 

[Nickie]

I would approach this problem by first converting the initial velocity of 30 mph to meters per second (m/s). This can be done by using the conversion factor 1 mph = 0.44704 m/s. Therefore, the initial velocity (Vo) is 13.41 m/s.

Next, I would use the equation V^2 = Vo^2 + 2aΔS to solve for the acceleration (a). We know that the final velocity (V) is 0 m/s (since the person comes to a complete stop) and the distance (ΔS) is 1 m. Plugging in these values, we get:

0^2 = (13.41)^2 + 2a(1)
0 = 179.88 + 2a
2a = -179.88
a = -89.94 m/s^2

Now that we have the acceleration, we can use the equation V = Vo + at to solve for the time (t). Again, we know that the final velocity (V) is 0 m/s and the initial velocity (Vo) is 13.41 m/s. Plugging in these values, we get:

0 = 13.41 + (-89.94)t
-13.41 = -89.94t
t = 0.149 s

Finally, we can use the equation ΔS = VoT + 1/2at^2 to solve for the average g-force. We know that the distance (ΔS) is 1 m, the initial velocity (Vo) is 13.41 m/s, and the time (t) is 0.149 s. Plugging in these values, we get:

1 = (13.41)(0.149) + 1/2(-89.94)(0.149)^2
1 = 2.00043 - 0.6731
0.6731 = 1.3269g
g = 0.508 g

Therefore, the average g-force experienced by someone abruptly coming to rest from an initial velocity of 30 mph in a distance of 1 m is approximately 0.508 g. This is equivalent to about half of the force of gravity experienced on Earth.