Answer to Question #105230 in Classical Mechanics for Jeff

Question

In the car that decelerates from 30 kph to rest over 15 cm are two 70 kg passengers. One wears a seatbelt and decelerates over 30 cm (with respect to the ground) (two significant figures). The other is not wearing a seatbelt and decelerates over 5 cm* (with respect to the ground). What are the magnitudes of the average external forces acting on the two passengers during their decelerations? (Careful with significant figures: do both answers have the same precision?)

a) With a seatbelt? _____ kN.

b) Without a seatbelt? _____ kN.

Please separate your answers with a comma. Example: a, b

This deceleration occurs when he strikes a part of the car, which is by then stationary. The 5 cm is the passenger's own crumple zone, and this is where the injuries occur.

Answer 1

Converting 30km/hr into m/s

u=25/3 m/s

v=0

We will use the expression $v^2 - u^2=2aS$

a) With the seat belt S=30 cm =0.3 m

$v^2 - u^2=2aS\\-\dfrac{625}{9}=2\times a\times 0.3\\a=-\dfrac{625}{5.4}ms^{-2}=-115.75ms^{-2}$

Force = ma = $70\times115.75=8.1kN$

b) Without the seat belt S=5 cm = 0.05 m

$v^2 - u^2=2aS\\-\dfrac{625}{9}=2\times a\times 0.05\\a=-\dfrac{625}{0.9}ms^{-2}=-694.4ms^{-2}$

Force = ma = $70 \times694.4 = 48.61 kN$