acceleration?

Question

A car traveling at 29 m/s (65 mi/h) runs into a bridge abutment after the driver falls asleep at the wheel.

If the driver is wearing a seat belt and comes to reat within a 1.0 m distance. what is his acceleration?

A passenger who isn't wearing a seat belt is thrown into the windshield and comes to a stop in a distance of 10.0 cm what is the acceleration of the passenger?

Okay, so we have to find two accelerations, what formulas do i use to find out what they are...

Answer 1

You just have to use one constant acceleration equation:

x = (v_f^2-v_0^2)/(2a)

We know x, v_f = 0 (person comes to a complete stop), and v_0 = 29 m/s. Now we can solve for 'a'

Passenger with seatbelt:
a = (v_f^2-v_0^2)/(2x) ------------>(0-29^2)/(2*1.0) = 420.5 m/s^2

Passenger without seatbelt:
a = (v_f^2-v_0^2)/(2x) ------------>(0-29^2)/(2*0.01) = 4.21x10^5 m/s^2

Hope this helps