Design of seabelt. Suppose that the car is stopped from 60mph and the driver (75

Question

Design of seabelt. Suppose that the car is stopped from 60mph and the driver (75kg) didn't push the brake. a. Seabelt is non-stretchable and produces a constant force to stop the driver in: i. 10cm ii. 20cm. Calculate the acceleration of the driver and the force that the seatbelt acts on the driver for above two conditions. b. Seatbelt is elastic (F_s = k*x). i. Please derive the expression of the distance needed to stop the driver as function of the elastic constant k. ii. If driver has to be stopped within 20cm, how strong the elastic constant (k) should be? c. If the driver didn't buckle up, and he is stopped within 3cm after hitting the wheel. What's the average acceleration and force?

Explanation / Answer

(a)

Time for the driver to hit the seatbelt: t=d/v

Force that the seatbelt exerts on the driver: F=m(vo-vt)/t

Acceleration of the driver: a=F/m

(i)

60mph = 27 m/s

Time for the driver to hit the seatbelt:

t=d/v = 0.1m/(27m/s) = 3.7ms

Force that the seatbelt exerts on the driver:

F=m(vo-vt)/t = 75kg(27m/s - 0m/s)/3.7ms = 547,297N

Acceleration of the driver: a=F/m = 547,297N/75kg = 7297.3 m/s^2

(ii)

Time for the driver to hit the seatbelt:

t=d/v = 0.2m/(27m/s) = 7.4ms

Force that the seatbelt exerts on the driver:

F=m(vo-vt)/t = 75kg(27m/s - 0m/s)/7.4ms = 273648N

Acceleration of the driver: a=F/m = 273648N/75kg = 3648.64 m/s^2

(b)

(i)

final velocity of the car = 0

the acceleration of the car is

a = v2 - vo2 /2x

fro the newtons second law

F = m*a

-k*x =  m*(v2 - vo2 /2x)

-k*x = m*(v2 - 0 /2x)

x = sqrt (m*vo2 / 2k)

(ii)

the elastic spring constant

k = mvo2 /2*x2

k = 75*27*27 / 2*0.20*0.20

k = 6.83 N/m

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