a 60kg woman is a passenger in a car that is involved in a crash. she is wearing a seat belt that stops her in 0.1s. if the car gad been travelling at 12m/s before the crash calculate the deceleration and the force on her breastbone due to the seat belt?
can you assume that v=om/s and u= 12m/s? so using v=u+at work out accel and then f=m.a for the force?
a = (0-12)/.1
F * t = change in m v
F * 0.1 = 60 (0-12)
a = change in velocity / change in time
and
F = change in momentum / change in time (Newton #2, F = d/dt (m v) = m a if m is constant)
Yes, I can assume that the initial velocity (u) is 12 m/s and the final velocity (v) is 0 m/s.
To calculate the deceleration (a), we can use the equation:
v = u + at
Rearranging the equation, we have:
a = (v - u) / t
Substituting the given values, we get:
a = (0 - 12) / 0.1
a = -120 m/s^2
Since the deceleration is negative, it means the car is slowing down.
Now, to calculate the force on the woman's breastbone due to the seat belt, we can use Newton's second law of motion, which states:
Force (F) = mass (m) × acceleration (a)
In this case, the mass (m) of the woman is given as 60 kg, and the acceleration (a) is -120 m/s^2.
Substituting the values into the equation, we get:
F = 60 kg × (-120 m/s^2)
F = -7,200 N
Therefore, the force on the woman's breastbone due to the seat belt is -7,200 N. The negative sign indicates that the force is acting in the opposite direction to the motion.