CATAGORY: Mechanics
QUESTION: it's not a textbook problem. just been driving me up the wall.
two cars, drivng towards each other (1-d). 25 m/s and 35 m/s respectively.
cars are 225m apart and hit the brakes, both decelerating at 4.0 m/s^2 ...
when do they collide and what's their relative velocities on impact?

Back to catalog.

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This is a good problem because it shows that sometimes a discovery is made
while working a problem that leads one to realize the situation is a
different one than originaly thought and that a different approach is
required.  

Choose an origin of coordinates to measure positions and velocities
relative too.

Take the car that initialy travels at 25m/s to be moving in the +
direction and has position 0 when the car starts to break.  The equation
of motion of the car as a function of time is

(1)     x = 0m + 25m/s t  - (.5)4.0m/s^2 t^2

The minus is because the car while moving in the + direction is slowing
down so the acceleration must be in the opposite sense of the velocity.
The equation of motion of the second car is then

(2)     x = 225m - 35m/s t + (.5)4.0m/s^2 t^2

Here when the braking starts the second car is 225meters from the origin
traveling toward the first car which means that the velocity is negative.
Since the second car is breaking its acceleration is in the opposite 
direction of the velocity so for this car the acceleration is +.

The two cars colide when they have the same positions. so

(3)   0m + 25m/s t  - (.5)4.0m/s^2 t^2 = 225m - 35m/s t + (.5)4.0m/s^2 t^2

Gathering together like terms on one side of the equation gives

(4)     0 = 225m - 35m/s t -25m/s t + (.5)4.0m/s^2 t^2 + (.5)4.0m/s^2 t^2

        0 = 225m - 60m/s t + 4.0m/s^2 t^2
        
from the quadratic formula we can solve for t.

(5)    t = [60 +/-sqrt( 60^2 -4(4.0)225)] /2(4)  seconds

         = 60/8 seconds 
         
         = 7.5 seconds.
         
the velocity of the two cars are 

(6)     V1 =  25m/s - 4.0m/s^2 t
        V2 = -35m/s + 4.0m/s^2 t 
        
Substitute 7.5 seconds into (6) to give
 
        V1 = -5 m/s
        V2 = -5 m/s
        
 AT THIS POINT WARNING FLAGS SHOULD GO UP!  THE FIRST CAR AS IT BRAKES 
 CAN NOT HAVE NEGATIVE VELOCITY.  THE CAR MUST HAVE STOPPED BEFORE
 THE SECOND CAR HITS IT!
 
With this knowledge we have to start the problem again.  The first
car stops at when V1 = 0. From (6) this happens at

(7)      t = (25m/s)/(4.0m/s^2) = 6.25 s

the first car is then stopped at position

(8)    x = 0m + 25m/s (6.25s)  - (.5)4.0m/s^2 (6.25s)^2

         = 78.125 m from the origin.
         
The second car collides with the first when it reaches this position.
Since the first car is stopped the relative velocity of the collision
will be the magnitude of the velocity of the second car.

(9)   78.125m = 225m - 35m/s t + (.5)4.0m/s^2 t^2 

         0 = 146.875m - 35m/s t + (.5)4.0m/s^2 t^2
         
Solving for t again using the quadratic formula gives

(10)   t = [35 +/-sqrt(35^2 -4(2.0)146.875)]/2(2.0)

         =  [35 +/- 7.071]/4 
         
         = [6.98 , 10.51]   seconds 
         
We take the shorter time interval because the second car does not collide 
twice! (Besides at the second time the car would have reversed direction
which is not possible for a `forward' moving breaking car)
The relative velocity at impact is then 

(11)  |V2| = |-35m/s + 4.0m/s^2 (6.98 s)| = |-7.08 m/s| = 7.08 m/s

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        *                                         *
        *  relative velocity at impact ~ 7 m/s    *
        *                                         *
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