a = change in velocity / change in time
= -16.7 / t
t = distance/ average speed
= .67/8.35 = .0802 second
so
a = -16.7 / .0802 = -208 agree
F = m a = 42 * -208 = -8736 Newtons AGREE
A car travelling at 60 km/h hits a bridge abutment. A passenger in the car moves forward a
distance of 67 cm (with respect to the road) while being brought to rest by an inflated air bag.
What force (assumed constant) acts on the passenger¡¦s upper torso, which has a mass of 42 kg?
I am learning physics by myself,and I found it a little bit difficult and there are no answers. I hope you can tell me if I am right.
My answer is :
v=final velocity; u=initial velocity
according to the question,
v=0,u=60*1000/3600 m/s = 16.7m/s
s=0.67m
a= (v^2 - u^2)/(2s)
a= (0 - 16.7^2)/(2*0.67)
a=-208 ms^-2
so the required F is
F=ma = 42 * 208 = 8736 N
is that correct?
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