How Do You Calculate Deceleration and Projectile Motion Problems?

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To calculate deceleration in motion problems, the formula x = x0 + v0t + (1/2)at² is used, where negative acceleration indicates deceleration. In the scenario with two trains, each decelerating at 1.0 m/s² from speeds of 72 km/h and 133 km/h, the braking distance can be determined using the equation vf² = vi² + 2ad, even without specified time. For the flowerpot problem, the total time of 0.35 seconds and the height of 2.10 m can be analyzed using kinematic equations, but additional information or assumptions may be necessary to solve for the maximum height above the window. Understanding the context and applying the correct formulas is crucial for solving these physics problems effectively.
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With motion problems, what formula do i use for deceleration of a particle if I am given v and a, and need to find distance?
 
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For constant acceleration:
x = x0 + v0t + (1/2)at2
 
So, to find braking distance for each train in this problem:

Two trains, one traveling at 72 km/h and the other at 133 km/h, are headed toward one another along a straight, level track. When they are 860 m apart, each engineer sees the other's train and applies the brakes. The brakes decelerate each train at the rate of 1.0 m/s2.

would I use negative acceleration?
 
and what do i do about the fact that time isn't specified in the question?
 
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html

look on the equations on this page...find any that might work?
 
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vf^2=vi^2+2ad

you are awesome
 
I have another question: How do I go about this one:

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.35 s, and the top-to-bottom height of the window is 2.10 m. How high above the window top did the flowerpot go?

I tried v=-gT+Vo and y=-.5gT^2+VoT
but I think I'm missing something in the way the question is phrased.
It gives total time, but no other numbers to plug in...
 
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