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We begin by stating the two points on the line which are known. They are ( t _{1} , v _{1} ) and ( t _{2} , v _{2} ). The numerical values of these points are (1.00 s, 17.00 m/s) and (2.00 s, 22.00 m/s) respectively.
Our knowledge of straight lines tells us that we can calculate the slope through a differencing operation
Next we enter the numerical values of the problem
We can therefore conclude that the acceleration is 5.00 m/s ^{2} . We can incorporate this value into the linear equation that relates velocity to time
Either of the data points can be used to solve for the intital velocity. Let us substitute the values associated with the first data point into the equation. We obtain
So we conclude that the initial velocity of the vehicle is 12.00 m/s and the constant acceleration of the car while it is situated on the on-ramp is 5.00 m/s ^{2} .
Suppose that we are presented with an electric circuit that contains a fixed voltage source ( v ), a variable source voltage ( v _{s} ) and a resistor ( R ). This situation is shown in Figure 3. In this figure, the variable source voltage is indicated by the circle and represented by the variable v _{s} . Its units are Volts. The current that flows through the circuit is indicated by the variable i and flows in the direction indicated by the arrow. The current has the units Amps. The fixed voltage source is represented by the constant v .
One of the most important laws of Physics that govern the behavior of electric circuits is Kirchoff’s Voltage Law. This law states the algebraic sum of the voltage drops experienced as one passes through a complete path through a circuit is equal to zero.
Application of Kirchoff’s Voltage Law to this circuit yields the equation
The terms of this equation may be arranged to produce the following equation
Let us consider the source voltage ( v _{s} ) as the dependent variable and the current ( i ) as the independent variable. Examination of the equation reveals that there is a linear relationship between v _{s} and i . For this linear equation, the value of the resistor ( R ) is the slope and the fixed voltage ( v ) represents the y -intercept.
Let us apply our knowledge of linear equations to solve a problem associated with this circuit.
Question: It is observed through measurement that the when the variable source voltage is 6.00 Volts, the current takes on the value of 1.00 Amp. When the variable source voltage is raised to 12.00 Volts, the current rises to a value of 1.50 Amps. Find the values for the resistance and the fixed voltage.
Solution: We can draw some insight into the solution of this problem by applying our knowledge of straight lines.
Let us begin by finding the value of the slope or equivalently the value of the resistance. We have two ordered points to consider (1.00 A, 6.00 V) and (1.50 A, 12.00 V). The slope of the line that connects these two points is
We recognize that the ratio (volts/amps) is equivalent to the unit (Ω). We make the substitution to yield
Earlier we stated that the slope of the line would be equal to the value of the resistance, so we have the following result
The next step in the solution is to solve for the value of the fixed voltage. Incorporation of the slope that was just found into the equation of the line yields the equation
Let us substitute the values 1.00 A and 6.00 V into this equation
So we conclude that the value of the fixed voltage is -6.00 V and that the value for the resistance is 12.0 Ω.
Knowling how to apply the knowledge of linear equations and straight lines is critical for students in engineering. In this module, we have seen how knowledge of linear equations can be used to solve engineering problems. Applications from the fields of fluid mechanics, the mechanics of motion and electric circuits have been presented. Other applications in engineering abound.
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