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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

m = v 2 v 1 t 2 t 1 size 12{m= { {v rSub { size 8{2} } - v rSub { size 8{1} } } over {t rSub { size 8{2} } - t rSub { size 8{1} } } } } {}

Next we enter the numerical values of the problem

m = 22 . 00 m / s 17 . 00 m / s 2 . 00 s 1 . 00 s size 12{m= { {"22" "." "00"`m/s - "17" "." "00"`m/s} over {2 "." "00"`s - 1 "." "00"`s} } } {}
m = 5 . 00 m / s 2 size 12{m=5 "." "00"`m/s rSup { size 8{2} } } {}

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

v ( t ) = ( 5 . 00 m / s 2 ) t + v 0 size 12{v \( t \) = \( 5 "." "00"`m/s rSup { size 8{2} } \) `t+v rSub { size 8{0} } } {}

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

17 . 00 m / s = ( 5 . 00 m / s 2 ) × ( 1 . 00 s ) + v 0 size 12{"17" "." "00"`m/s= \( 5 "." "00"`m/s rSup { size 8{2} } \) times \( 1 "." "00"`s \) +v rSub { size 8{0} } } {}
17 . 00 m / s = 5 . 00 m / s + v 0 size 12{"17" "." "00"`m/s=5 "." "00"`m/s+v rSub { size 8{0} } } {}
v 0 = 12 . 00 m / s size 12{v rSub { size 8{0} } ="12" "." "00"`m/s} {}

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 .

Electric circuits – variable source voltage

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 .

Electrical circuit with a variable source voltage.

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

v s + R i + v = 0 size 12{ - v rSub { size 8{s} } +R`i+v=0} {}

The terms of this equation may be arranged to produce the following equation

v s = R i + v size 12{v rSub { size 8{s} } =R`i+v} {}

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

m = v 2 v 1 i 2 i 1 size 12{m= { {v rSub { size 8{2} } - v rSub { size 8{1} } } over {i rSub { size 8{2} } - i rSub { size 8{1} } } } } {}
m = 12 . 00 V 6 . 00 V 1 . 50 A 1 . 00 A size 12{m= { {"12" "." "00"`V - 6 "." "00"`V} over {1 "." "50"`A - 1 "." "00"`A} } } {}
m = 12 . 0 ( V / A ) size 12{m="12" "." 0` \( V/A \) } {}

We recognize that the ratio (volts/amps) is equivalent to the unit (Ω). We make the substitution to yield

m = 12 . 0 Ω size 12{m="12" "." 0` %OMEGA } {}

Earlier we stated that the slope of the line would be equal to the value of the resistance, so we have the following result

R = 12 . 0 Ω size 12{R="12" "." 0` %OMEGA } {}

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

v s = 12 . 0 ( V / A ) × i + v size 12{v rSub { size 8{s} } ="12" "." 0` \( V/A \) ` times i`+v} {}

Let us substitute the values 1.00 A and 6.00 V into this equation

6 . 00 V = 12 . 00 ( V / A ) × 1 . 00 A + v size 12{6 "." "00"`V="12" "." "00"` \( V/A \) times 1 "." "00"`A+v} {}
6 . 00 V = 12 . 00 V + v size 12{6 "." "00"`V="12" "." "00"`V+v} {}
v = 6 . 00 V size 12{v= - 6 "." "00"`V} {}

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.


  1. Water flows through a piping system that consists of two pipes that are joined together. The cross-sectional area of the first pipe is 10.00 cm 2 , while that of the second pipe is 1.25 cm 2 . If the velocity of the water in the first pipe is known to be 5 cm/s, then what is the velocity of the water in the second pipe. Assume that the continuity equation holds.
  2. A vehicle is traveling through a neighborhood at an initial velocity ( v 0 ). The driver of the vehicle notices a child who runs out into the street in front of her car. She applies her brakes to reduce the speed of her car and eventually stops. The velocity of her car obeys the linear relationship v ( t ) = v 0 + a t size 12{v \( t \) =v rSub { size 8{0} } +a`t} {} Determine the initial velocity and the acceleration ( a ) if the velocity is known to be 30 m/s at t = 0.50 s and the velocity is 3 m/s at the time t = 1.25 s. Also calculate the total time that it will take the vehicle to stop after the driver applies her brakes.
  3. Consider the circuit depicted in Figure 3. The following two facts are known. When the variable voltage is set to 9 Volts, the current is 100 mA. When the variable voltage is sent to 18 Volts, the current is 1.20 A. What are the values for R and v ?

Questions & Answers

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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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