Final answer:
The force on a 61-kg person accelerating at 30 g's is 17,934 N. If brought to rest from 92 km/hr, the distance traveled is 2.21 m.
Step-by-step explanation:
To calculate the force on a 61-kg person experiencing deceleration at a rate of 30 g's, we first need to express the deceleration in standard units. Since 1 g is equivalent to 9.8 m/s2, a deceleration of 30 g's is 30 × 9.8 m/s2 = 294 m/s2. Using Newton's second law (F=ma), the force exerted on the person is F = m × a = 61 kg × 294 m/s2 = 17,934 N.
To find the distance traveled before coming to rest from 92 km/hr, which is equivalent to 25.56 m/s, we can use the equation derived from the kinematic equations: v2 = u2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (25.56 m/s), a is the acceleration (-294 m/s2, negative because it's deceleration), and s is the distance. Rearranging for s gives s = (v2 - u2)/(2a). Plugging in the values gives s = (0 - (25.56 m/s)2)/(2 × -294 m/s2) = 2.21 m.