1. vehicle stopping distances

I am given the speed in km/h (s) and total stopping distance in meters (t)
Using the rule t = as^2 find values from 100 to 150.
I looked in my pre-calc book and there is an equation for stopping distance which is d = v +(v^2/20) and this confused me. I did not get the same results, is it because it is in mpr and not kmr.
Thank You,
Keith

2. Originally Posted by keith
I am given the speed in km/h (s) and total stopping distance in meters (t)
Using the rule t = as^2 find values from 100 to 150.
I looked in my pre-calc book and there is an equation for stopping distance which is d = v +(v^2/20) and this confused me. I did not get the same results, is it because it is in mpr and not kmr.
Thank You,
Keith
Whaaaaat?

If you have an object undergoing constant acceleration (in the direction opposite to the velocity, ie a deceleration) and an initial velocity v0 then the stopping distance may be calculated as the position from the starting point where the velocity goes to 0. ie.

d = -v0^2/(2a)
(The negative sign appears because I have tacitly assumed a positive direction as the direction of the initial velocity, and hence the acceleration, a, is negative.)

Neither of the equations:
t = as^2
d = v +(v^2/20)
come out in the correct units. I would treat them as highly suspect. (Or more likely as applications to specific problems.)

-Dan

3. vehicle stopping distances

This question was one part of a problem in my math education class and it does not make any sense the way it is written.
Thank You for looking it over. I will ask the professor when I get to class tonight.
Thank you,
Keith