Question #154175

In the event of a collision between vehicles, the severity of injuries sustained by drivers can be reduced with an air bag. In a car initially traveling at +100 km / h, an air bag immobilizes a driver with a mass of 62 kg in 0.090 s. The magnitude of the average force exerted by the air bag on the driver is?


1
Expert's answer
2021-01-10T18:29:17-0500

Let's first convert km/hr to m/s:


vi=(100 kmh)(1000 m1 km)(3600 1 h3600 s)=27.77 ms.v_i=(100\ \dfrac{km}{h})\cdot(\dfrac{1000\ m}{1\ km})\cdot(3600\ \dfrac{1\ h}{3600\ s})=27.77\ \dfrac{m}{s}.

We can find the average force exerted by the air bag on the driver from the impulse-momentum change equation:


FavgΔt=mΔv,F_{avg}\Delta t=m\Delta v,FavgΔt=m(vfvi),F_{avg}\Delta t=m(v_f-v_i),Favg=m(vfvi)Δt,F_{avg}=\dfrac{m(v_f-v_i)}{\Delta t},Favg=62 kg(0 ms27.77 ms)0.090 s=1.9104 N.F_{avg}=\dfrac{62\ kg\cdot(0\ \dfrac{m}{s}-27.77\ \dfrac{m}{s})}{0.090\ s}=-1.9\cdot10^4\ N.

The magnitude of the average force exerted by the air bag on the driver equals 1.9104 N.1.9\cdot10^4\ N. The sign minus means that the average force exerted by the air bag on the driver directed in the opposite direction to the motion of the car.

Answer:

Favg=1.9104 N.F_{avg}=1.9\cdot10^4\ N.


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