A 60-kg crash test dummy moving at 13.4 m/s (30 mi/hr) stops during a collision

Question

A 60-kg crash test dummy moving at 13.4 m/s (30 mi/hr) stops during a collision in a distance of 0.65 m. Estimate the average force that the air bag and seat belt exert on the dummy.

Answer for acceleration is -138 m/s/s and answer for force is 8300, but I don't understand how.

I'm only allowed to use equations:

1. v = Dx/Dt  

2.x = vt + x0

Constant Acceleration

3. a = Dv/Dt

4. v = v0 + at

5. x = x0 + v0t + ½ at2

6. x – x0 = (v2 - v02)/2a

7.x – x0 = ½ (v + v0)t

Forces

8. a = F / m     (or F = ma)

9. FE on O = mog      (*g = 9.8 N/kg 10 N/kg)

10. F1 on 2 = - F2 on 1

Explanation / Answer

mass = 60 kg

initial velocity vo = 13.4 m/s

final velocity v = 0

distance travelled = x - xo = 0.65 m

from 6

x – x0 = (v^2 - v0^2)/2a

0.65 = (0^2-13.4^2)/(2*a)

a = -138 m/s/s

from (8)

F = m*a

F = 60*138 = 8300 N

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