Simple motorboat kinematics problem

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SUMMARY

The discussion centers on solving a motorboat kinematics problem involving uniform acceleration. The motorboat slows from 75 km/h to 40 km/h over a distance of 50 m, and the key equation used is v² = v₀² + 2a(x - x₀) to find the acceleration. The user seeks to understand the derivation of this equation without relying solely on memorization. A step-by-step integration approach is provided, leading to the elimination of time to derive the kinematics formula.

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maxbashi
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I'm stuck doing this problem in my first year mechanics class. I feel sort of dumb for not being able to get it. The teacher wants us to just memorize the formulas but I really really really don't want to do that - I'm trying to just do it using the basics and calculus. I can obviously do this problem by plugging in numbers, but I'm looking for a way to do it from scratch.


Homework Statement


A motorboat traveling on a straight course slows dow uniformly from 75 km/h to 40 km/h in a distance of 50 m. What is the magnitude of its acceleration?


Homework Equations


The one that you're "supposed" to used to get the right answer is v^2=vo^2+2a(x-xo).
I'm just trying to figure out why one would derive it this way - why the square speed? Obviously it's dimensionally needed to get m/s^2 for the answer... but I'm just looking for any explanation I can get.

The Attempt at a Solution


several approaches I've tried...
v(t)=(75-at)km/h - but don't know how to find what t would be, since only the distance (50m) is given

average acceleration = (75-40)/t... again, the problem is that I can't find t.

I would appreciate any help you guys could give me. And if I can provide any more information about my thoughts so far let me know. Thanks
 
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From scratch eh? Okay:

You start with the acceleration as a function of time:

a(t) = a = const. (1)​

Integrate this once with respect to time to get the velocity as a function of time :

v(t) = v0 + at (2)​

Here, v0 is the constant of integration, and it is obviously equal to the initial velocity (which you'll see if you set t = 0). Integrate this function with respect to time to get the position function:

x(t) = x0 + v0t + (1/2)at2 (3)​

Now you've derived all but one of the kinematics formulae. The last one makes no explicit mention of time. To eliminate t from the equations, solve for t in (2):

t = (v - v0) / a​

Substitute this into (3), rearrange, and you will end up with the result that:

v2 - v02 = 2a(x - x0) (4)​

An easier way to derive (4) is to use the work-energy theorem. As soon as your prof teaches you what that is, try it!
 
Thank you so much! Eliminating t by setting it equal to (v-vo)/a was just what I was missing. Thanks!
 

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