question!

• Jun 27th 2006, 02:14 PM
LLELLA
question!
In an emergency stop to avoid an accident, a shoulder-strap seat belt holds a 60-kg passenger firmly in place. If the car were initially traveling at 90 km/h and came to a stop in 5.5 s along a straight, level road, what was the average force applied to the passenger by the seatbelt?
• Jun 27th 2006, 02:36 PM
Jameson
First thing is convert all units to standard ones.

$\frac{90 km}{h} \times \frac{1000m}{1km} \times \frac{1h}{60min} \times \frac{1min}{60sec}=\frac{25m}{sec}$

So if $F_{net}=ma$ then we need to solve for acceleration since we have the mass. Use the equation $v_{f}=v_{0}+at$. Plugging in your values shows $0=25+5.5a$ and a=4.55 m/s^2.

This is the average acceleration, thus the average force is 4.55*60=273N