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In: Physics

Using kinematic relation s = v_0t +(1/2)at^2 , I got the 2 equations: 172.8m = 5.682V_0...

Using kinematic relation s = v_0t +(1/2)at^2 , I got the 2 equations: 172.8m = 5.682V_0 + 16.143a and 140.3m = 4.930V_0 + 12.152a. How do I get Initial velocity and acceleration out of that? m=meters and t is seconds

Solutions

Expert Solution

172.8 = 5.682V_0 + 16.143a



140.3 = 4.930V_0 + 12.152a



so these are two linear equation


let V_0 = x and a = y


so.

5.682x + 16.143y = 172.8 --- i)


4.930x + 12.152y = 140.3 --- ii)



The goal here is to solve

5.682x + 16.143y=172.8 and

4.930x + 12.152y=140.3 for the variables x and y.

First, let's work on your first equation,5.682x + 16.143y=172.8


This means, see if it can be simplified at all before attempting to solve it.

Multiply x and 5.682


Multiply x and 1



The x just gets copied along.


The answer is x


x


5.682*x evaluates to 5.682x


Multiply y and 16.143


Multiply y and 1



The y just gets copied along.


The answer is y


y


16.143*y evaluates to 16.143y


5.682*x+16.143*y evaluates to 5.682x+16.143y


So, all-in-all, your first equation can be written as: 5.682x+16.143y = 172.8


Now, let's work on your second equation,4.930x + 12.152y=140.3 Multiply x and 4.93


Multiply x and 1



The x just gets copied along.


The answer is x


x


4.930*x evaluates to 4.93x


Multiply y and 12.152


Multiply y and 1



The y just gets copied along.


The answer is y


y


12.152*y evaluates to 12.152y


4.930*x+12.152*y evaluates to 4.93x+12.152y


So, all-in-all, your second equation can be written as: 4.93x+12.152y = 140.3


After this initial survey of the equations, the system of equations we'll set out to solve is:

5.682x+16.143y = 172.8 and 4.93x+12.152y = 140.3

Let's start by solving 5.682x+16.143y = 172.8 for the variable x.

Move the 16.143y to the right hand side by subtracting 16.143y from both sides, like this:

From the left hand side:

16.143y - 16.143y = 0


The answer is 5.682x


From the right hand side:

The answer is 172.8-16.143y



Now, the equation reads:

5.682x = 172.8-16.143y

To isolate the x, we have to divide both sides of the equation by the other variables

around the x on the left side of the equation.


The last step is to divide both sides of the equation by 5.682 like this:

To divide x by 1



The x just gets copied along in the numerator.


The answer is x


5.682x


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