How Is Force Calculated When a Student Hits a Dashboard?

[gneill]

Apparently I'm retarded but I just don't understand how you used this formula the way you did.

d = Vi x t + (1/2) a x t^2

As far as I know, this formula is for calculating the distance that an object travels over a specified time interval when you are given how long it takes, the initial velocity, and the acceleration.

Why do you sub in Vi x t as being the distance from the dashboard? Sure, the acceleration is -13.89 m/s^2, but isn't the d on the left side of the equation 0.65m?

The complete general expression is

d(t) = d_o + Vi x t + (1/2) a x t²

d_o is the initial displacement.

d(t) is the separation at time t.

In this case, in the frame of reference of the moving car and student at the instant the crash begins, Vi = 0; they are moving at the same rate in the ground frame of reference, so in the student/car frame their relative velocity Vi is zero. When you then put in the negative value for the acceleration, you obtain the formula I gave.

 

[Oriako]

Wow. Why do they remove that d sub-zero term on my formula sheet then!? WTF. It should be: ∆d = Vi x t + (1/2) a x t^2

Thank you so much for all the help you've given me. This problem has been bothering me for days.