posted by Courtney .
there's three parts, i keep getting these wrong please help.
A set of crash tests consists of running a test car moving at a speed of 11.8 m/s (26.4 mi/hr) into a solid wall. Strapped securely in an advanced seat belt system, a 59.0 kg (130 lbs) dummy is found to move a distance of 0.660 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.
Using the direction of motion as the positive direction, calculate the average acceleration of the dummy during that time (in g's, with 1g = 9.81m/s2).
In a different car, the distance the dummy moves while being stopped is reduced from 0.660 m to 0.210 m,
calculate the average force on the dummy as that car stops.
(Average belt force)x (distnce moved)
= initial kinetic energy of dummy
= (1/2)*59 kg* (11.8 m/s)^2 = 4108 J
Avg. Force = 4108/.66 = 6224 Newtons
Acceleration is backwards (negative sign) and equals the speed change divided by time intervsl
sqrt (-2aX) = V
a = -V^2/(2X) = -105.5 m/s^2
= -10.75 g's
Repeat the same method for the second case. The force must be larger to stop in a reduced distance