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A car travelling at 72 km/h hits a wall and comes to a stop. The average deceleration experienced by the car in this duration is 6g (use g=10 m/s 2 ). The belted passenger, weighing 1000 N, is restrained by the belt that exerts a constant force of 4 kN. The steering wheel of the car is 250 mm from the occupant. Will the seat belt provide adequate protection to the occupant against hitting the steering wheel? If not, with what velocity will the occupant hit the steering wheel? (4+3) marks Had the distance between the occupant and the steering wheel been 150 mm, and an air bag had to be used for supplemental protection, at what time should the bag be triggered if the gas to fill the bag of volume is produced by chemical reaction at a rate of 2.5 kl/sec. Assume that the bag is spherical in shape and has a radius of 250 mm.

 
A car travelling at 72 km/h hits a wall and comes to a stop.  The average deceleration experienced by the car in this duration is 6g (use g=10 m/s2).  The belted passenger, weighing 1000 N, is restrained by the belt that exerts a constant force of 4 kN.  The steering wheel of the car is 250 mm from the occupant.
  1. Will the seat belt provide adequate protection to the occupant against hitting the steering wheel? If not, with what velocity will the occupant hit the steering wheel?
(4+3) marks
  1. Had the distance between the occupant and the steering wheel been 150 mm, and an air bag had to be used for supplemental protection, at what time should the bag be triggered if the gas to fill the bag of volume is produced by chemical reaction at a rate of 2.5 kl/sec.  Assume that the bag is spherical in shape and has a radius of 250 mm.

Grade:12th pass

1 Answers

Shivam Chopra
45 Points
9 years ago
The mass of person is 100 kg.   So, the force on man is 60*100 = 6000N
But , restrained force of belt is 4000N
So, resulting force 6000-4000 = 2000N
So, he will strike the steering wheel at the acceleration of 20ms-2
Now, s=distance=0.25m, acc=20ms-2
So, using v2-u2 = 2as,  u=0 because man is at rest
v2=2*20*0.25= 10     =>v = 3.16ms-1 . He will strike wheel at v = 3.16ms-1
 
Now, if s=0.15m, a=20ms-2   ;  so v2=2*20*0.15 = 6  =>v = 2.44ms-1
Now, using v = u+a*t    putting u = 0;
 t = v/a = 2.44/20 = 0.122 sec
Also, radius of bag = 25cm;
Volume = 4/3 πr3  = 4/3*3.14*25*25*25 cm3 = 65416 cm3 appx = 65.4 litre
Rate of filling = 2500 litre sec-1
So, time taken to fill bag = 65.4/2500 = 0.02616 sec
So, the bag should open 0.12 – 0.02616 = 0.09384 seconds after colloision
Please approve if you like it
Thanks

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