- #1
busbus
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Homework Statement
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)
Homework 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.
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...