Homework Help: Distance it takes to stop HELP

1. Sep 17, 2008

jo3jo3520

The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the driver's reaction time of 0.50 s.

(a) What is the minimum stopping distance for the same car traveling at a speed of 50 m/s?

Vf^2 = Vi^2 +2ax

So i would be solving for X here right? so the car is not accelerating do i plug in 0 for a?

This is what i have so far

0=50^2+2(a)(x) where does the .5 reaction time come into play?

2. Sep 17, 2008

LowlyPion

You're right that the .5sec reaction time needs to be accounted for.

But you need to first account for it from the first measurements.

The Total distance traveled to stopping included .5*(30m/s). Then the total change in velocity occurred over that smaller distance - not including reaction time.

Now for the 50m/s case how long to stop if you use the deceleration from the first measured stop? From that distance you then need to add the (50m/s)*.5sec to get your total distance.

3. Sep 17, 2008

jo3jo3520

ok so .5*30 = 15 but im confused on what to do with that? you said "Then the total change in velocity occurred over that smaller distance - not including reaction time." ?

4. Sep 17, 2008

jo3jo3520

am i supposed to do 30/15? and where in the equation do i plug this number in?

5. Sep 17, 2008

jo3jo3520

help anyone?

6. Sep 17, 2008

LowlyPion

So the car deceleration alone was only over 45m and not 60m then wasn't it? Now figure the deceleration performance of just the car.

You have Velocity and distance, so find acceleration "a" that gives you that deceleration over 45m.

7. Sep 17, 2008

jo3jo3520

Either Im just really dumb or not understanding but i give up. thank you for your help