# 2 Collision Avoidance Problems: Constant Deceleration given (a) time and (b) distance

1. Sep 3, 2010

### busbus

1. The problem statement, all variables and given/known data

I have to write a program that will calculate the distance and/or deceleration needed to stop a car following another. My teacher gave us an example:

(1) When driving at a 60 mph, one covers 44 feet in 0.5 sec.
(2) At 1 g deceleration, one would need 121 feet to come to a stop.
(3) At 2 second tailing, with the 0.5-second reaction time, one would need to decelerate at about 15.3 ft/sec² (0.48 g)
(4) At 11' tailing, deceleration would be 44 ft/sec² (1.38 g)

2. Relevant equations

The first two are easy to understand.

Constant velocity formula is x = v t (where x = distance traveled, v = initial velocity, and t = time).

(This also means that your velocity is 88 feet per second if you are going 60 MPH)

I also know that the deceleration formula is v² = 2 a x (where v = velocity, a = deceleration, and x = distance traveled).

those answer the first two questions.

The second two are killing me because I cannot determine how he arrived at his answers. I realize that the 2-second tailing would be equivalent to 242 feet (at 60 mph) and the 0.5 reaction time would mean I lost 44 feet before I started to decelerate. The 11-foot tailing example has me totally baffled.

3. The attempt at a solution

I DON'T KNOW WHERE TO START!!!! I think I am thinking about it too much. For (3), I am thinking I need to remove the reaction time distance (44 feet) from my cushion distance (242 feet). That means I started constant deceleration at 198 feet. The guy in front of me will stop at 121 feet. I also know I need to stop in enough time that I do not touch him, so i am subtracting 1 from something. Do I need to stop in, say, 197 feet (242' cushion - 44' reaction time - 1' so I don't crack him) = 197' to come to a stop. But I don't think that is right.

From above, a = v²/2x = (88fps)²/2(198) = 7744 / 396 = 19.55555. I am WAY off!!!

(4) is impossible for my brain right now....

2. Sep 3, 2010

### busbus

Re: 2 Collision Avoidance Problems: Constant Deceleration given (a) time and (b) dist

Wait a minute....I think I need to do something with the stopping distance of the guy in front of me...

I am going 88 feet per second. I am behind him at two seconds. That would be 2 * 88 = 196 feet. (NOT 242 feet!!!!)

It takes me 0.5 seconds to react, so I lose 44 feet right off the bat.

So that means I have 196 - 44 = 152 feet.

The guy in front stops is 121 feet.

Of course, if i decelerate exactly as quickly as him, I would also stop in 121 feet. That means I would be 152-121 = 31 feet behind him...but that is not the answer.

I need the MINIMUM deceleration.

Deceleration = Velocity^2 / (2 * Distance)

Deceleration = 88^2 / (2 * 31) = 7744 / 62 = 124.9999 ARRGGGHHHH!!!!!!!!

I am missing something somewhere. I just need a little guidance.

3. Sep 3, 2010

### busbus

Re: 2 Collision Avoidance Problems: Constant Deceleration given (a) time and (b) dist

Okay, maybe I got NUMBER 3:

(1) When driving at a 60 mph, one covers 44 feet in 0.5 sec. (88 feet in 1.0 sec)
(2) At 1 g deceleration, one would need 121 feet to come to a stop.
(3) At 2 second tailing, with the 0.5-second reaction time, one would need to decelerate at about 15.3 ft/sec² (0.48 g)
(4) At 11' tailing, deceleration would be 44 ft/sec² (1.38 g)

The deceleration formula is v² = 2 a x (where v = velocity, a = deceleration, and x = distance traveled).

I need to find DECELERATION, so that formula is a = v² / 2x.

Velocity for a car going 60 MPH is 88 feet per minute.
I am also going 60 MPH but I need to stop, say, one foot behind the guy in front.
The distance I need to stop is 242 feet but I need to add 1, so I need 243 feet

v = 88
x = 243 feet

a = v² / 2x
a = (88^2) / (2 *243)
a = 7744 / 486
a = 15.93416

I am getting close....

His is something like 7744 / ((242*2) + 22) = 7744 / (484 + 22) = 7744 / 506 = 15.30435

Where does that 22 come from? Reaction time is 0.5 seconds. That means it took me 44 seconds before I started to brake. Why would I divide 44 by 2??

I think I am way off. I am trying but my logic is all wrong.

Thanks.
busbus

4. Sep 3, 2010

### busbus

Re: 2 Collision Avoidance Problems: Constant Deceleration given (a) time and (b) dist

I figured it out!!!!! Thanks for letting me spout off.

Basically:

Determine how much deceleration you need to stop if you were seconds behind the guy in front of you and it took you 0.5 seconds before you put on the brake and started to decelerate

Cushion_Time 2 seconds
Cushion_Distance 176 feet
Reaction_Time 0.5 seconds
Reaction_Time_Distance 44 feet
Distance 121 feet (Stopping distance)

I need to first calculate the Total_Stopping_Distance = (Cushion_Distance + Distance) - Reaction_Time_Distance:

Cushion_Distance 176 feet
Distance 121 feet
Reaction_Time_Distance 44 feet
Total_Stopping_Distance 253 feet

Next, use this formula to get DECELERATION:
Deceleration = Velocity^2 / (2 * Distance)

Velocity 88 Feet_Per_Second
Total_Stopping_Distance 253 feet
DECELERATION 15.30434783
g 0.47826087

The formula is:
DECELERATION = 88^2 / (2 * 253) = 7744 / 506 = 15.30435.
g = DECELERATION / 32 = 0.47826087

---------------------------

Determine how much deceleration you need to stop if you were 11 feet behind the guy in front of you and it took you 0.5 seconds before you put on the brake and started to decelerate
Cushion_Distance 11 feet
Reaction_Time 0.5 seconds
Reaction_Time_Distance 44 feet = (Reaction_Time * Feet_Per_Second)
Distance 121 feet

Calculate the Total_Stopping_Distance =
(Cushion_Distance + Distance) - Reaction_Time_Distance
Cushion_Distance 11 feet
Distance 121 feet
Reaction_Time_Distance 44 feet
Total_Stopping_Distance 88 feet

Deceleration = Velocity 2 / (2 * Distance)

Velocity 88 Feet_Per_Second
Total_Stopping_Distance 88 feet
DECELERATION 44
g 1.375

DECELERATION = 88^2 / (2 * 88) = 7744 / 176 = 44
g = DECELERATION / 32

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So I got everything. I'm sorry if I did not explain myself right in here! I get the feeling I did a bad job explaining.

busbus