Solving Speeding: Will You Collide?

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The discussion focuses on a physics problem involving two cars: one speeding at 85 km/h and another traveling at the legal limit of 60 km/h. The speeder decelerates at 4.2 m/s² while the other car maintains a constant speed. The position equations for both cars are established using the formula X = X(0) + V(0)*t + 1/2*a*t². The participants aim to determine if a collision occurs and the distance at closest approach by solving for time and position.

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You're speeding at 85 km/h when you notice that you're only 10m behind the car in front of you, which is moving at the legal speed limit of 60 km/h. You slam on your brakes, and your car decelerates at 4.2 m/s^2. Assuming the car in front of you continues at constant speed, will you collide? if so, at what relative speed? if not what will the distance between the cars at their closest approach?

First I'm going to use the position formula (X= X(0) + V(0)*t + 1/2*a*t^2) to find the position of the two cars.

for the speeder(i'll car this car1):
V(0) = 85 Km/h or 23.6 m/s
a = -4.2 (decelerates so it's negative)
X(0) = 0
and time is unknown.

now for the second car:
a = 0 (since acceleration is constant)
x(0) = x_car1 + 10m (position of first car plus 10m, is this right?)
V(0) = 60km/h or 16.67 m/s

now i will plug in the known values:
(this is for car1) 10 m= 23.6*t + 1/2*-4.2*t^2
(this is for car2) 0 = (10m) + 16.67*t + 1/2(-4.2)*t^2

if i solve for time, what will that tell me? i mean will it tell me that it will take __ secs to get to the second car? did i set up everything correctly so far?
 
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Okay,the setup is incomplete.Sove for the time from the velocity equation and then plug in the first equation and in the second.If the 2 "x" coincide or the one for the decelerating car is bigger,then the 2 cars collide.

Daniel.
 
It's easier to work the problem out in a reference frame attached to the car moving at a fixed speed. :-)
 

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