# Homework Help: Physics Kinematics problems

1. Jul 28, 2008

### cerulescence

kinematics and free fall problems

The problem statement, all variables and given/known data
1). Suppose a car is traveling at 12.0 m/s, and the driver sees a traffic light turn red. After 0.510s have elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 6.20 m/s^2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light?

2). A cement block accidentally falls from rest from the ledge of a 53.0m high building. When the block is 14.0m above the ground, a 2.00m tall man looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

Relevant equations
1). I tried to use the equation: x = v0t + 1/2 at^2.
2). see below

The attempt at a solution
1). I tried saying that a = -6.20 m/s^2, the initial velocity v0 = 12, and that time = 0.510s, but I didn't get the right answer (17.7m).
2). a = -9.80 while initial velocity v0 = 0. However, I'm not sure exactly how I should go around calculating displacement. After I find displacement though, should I use the equation x = v0t + 1/2 at^2 to find the time?

Any pointers would be greatly appreciated!

Last edited: Jul 28, 2008
2. Jul 28, 2008

### Kurdt

Staff Emeritus
For part one there are two different scenarios you have to consider. The first is the reaction time. During this time the car is travelling at a constant speed and you need to find how far it goes. The second the car is slowing to a stop so you need to find out how far the car travels when coming to a stop and add it to the reaction time distance.

For the second question one can simply work out the time taken to fall to 14m above the ground and then 2m above the ground and find the difference.

3. Sep 30, 2009

### jkjkhardcore

hey thnx pf mentor that really helped, i was so stuck on that one. to answer op, what i did was get the distance from for the reaction time. I did D=TV .510secx12m/s= 6.12 Meters then i did the part with the deceleration Vf=0 Vo=12m/s a=-6.2m/s^2 time=? distance we need. Looking at the 4 forumulas only one applied Vf^2=Vo^2+2ax do algebra get x alone u get [Vf^2-Vo^2] divided by 2a i got 11.61Meters
then add them together 11.61meters + 6.12 meters is what?
I am not sure if i did that right though haha.

4. Sep 30, 2009

### jkjkhardcore

the second one i can't solve for time or Vf (final velocity) those are the two variables we have:
@ 51 Meters the guy's doomed, but at 39Meters he sees it.
so x=39 the Original Velocity = Vo=0, the acceleration is 9.8m/per sec per sec. due to gravity. sooo... now what? with the formulas..:
Vf=Vo+at... we don't have Vf and we dont have t.. we would like to find T.
Vf^2=Vo^2+2ax... we don't have Vf... maybe we find VF?
i got 27.6m/s for Vf now lets try to find time and i get 2.82 seconds... but now i just realized that finding the time from dropping it to where the man sees it is almost point less right? >.< but the Velocity where the man sees it is important. that's our new Vo.
sooo:
Vo=27.64778472
Vf=???
X(distance)=14-2 or 12meters
a=-9.8 (gravity
t=we don't know, funny how this looks exactly the same as before...
well solving for Vf i got 14.4066
i got 1.35sec.
soo lets find time:
Vf=Vo+at