Aperson in a car during a sudden stop can experience potentially serious chest injuries if the combined force exerted by the seat belt and shoulder strap exceeds 16,000 n. assume the mass of the passenger is 80 kg and the initial speed of the car is 16 m/s. describe what it would take to avoid injury.

minimum time interval to stop = 0.08 seconds

minimum stopping distance  = 0.64 m

Explanation:

maximum force (F) = 16,000 N

mass (m) = 80 kg

initial velocity (U) =  16 m/s

what it would take for the passenger to avoid in this case refers to how long it would take the vehicle to come to a full stop and the stopping distance it would also take to come to a full stop. Therefore we are to find the time (t) and the distance (s)

from the impulse momentum equation,

impulse = change in momentum

Ft = m(V-U)   (V-U is the change in velocity Δv)

where V is the final velocity = 0

and t = time

16000 x t = 80 (0 - 16)

16000t = -1,280 (he negative sign tell us there is a decrease in momentum, so we would not be using it further)

t = 0.08 seconds   ( this is also the difference between the initial time when the vehicle started to come to a stop and the final time when it came to a full stop)

assuming the acceleration is constant, the stopping distance (s) would be given by the kinetic relation

change in distance (Δs) = \frac{(ΔV) x (Δt)}{2}

(Δ refers to change, that is final value - initial value)

Δs =  \frac{16 x 0.08}{2}

Δs = 0.64 m