Answer to Question #105552 in Classical Mechanics for Charmie Llenarez

Question

In the car that decelerates from 30 \text{ kph}30 kph to rest over 15 \text{ cm}15 cm are two 70 \text{ kg}70 kg passengers. One wears a seatbelt and decelerates over 30 \text{ cm}30 cm (with respect to the ground) (two significant figures). The other is not wearing a seatbelt and decelerates over 5 \text{ cm}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? _____ \text{kN}kN.

b) Without a seatbelt? _____ \text{kN}kN.

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

Answer 1

First, convert kph to m/s:

$$v=30\cdot\frac{1000}{3600}=8.3\text{ m/s}.$$

The braking deceleration of the first passenger:

$$a_1=\frac{v^2}{2d_1},\\ \space\\ F_1=m_1a_1=m\frac{v^2}{2d_1}=8100\text{ N}.$$

No matter how long it took for the car to stop, the first passenger passed 30 cm relative to the ground to the full stop because his seatbelt stretched. It is interesting how the second passenger could go 5 cm relative to the ground while the car came to rest across 15 cm, perhaps that side of the car collided with something.

The second passenger stops with the deceleration

$$a_2=\frac{v^2}{2d_2},\\ F_2=m_2a_2=m\frac{v^2}{2d_2}=49000\text{ N}.$$

Therefore, the answer is 8.1 kN and 49 kN.