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?

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

Your calculations are correct! Well done!

To solve this problem, you correctly used the formula for acceleration (a) which is given by:

a = (v^2 - u^2)/(2s)

Where:
- v is the final velocity (in this case, v = 0 m/s, as the passenger comes to rest)
- u is the initial velocity of the car (u = 60 km/h converted to m/s)
- s is the distance traveled by the passenger (s = 0.67 m)

After calculating the acceleration, you used Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a):

F = ma

You correctly substituted the values of mass (m = 42 kg) and acceleration (a = -208 m/s^2) into the formula to find the required force.

So yes, your answer of 8736 N is correct! Keep up the good work!