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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.

Summary

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.

Exercises

  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

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
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hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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y=10×
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Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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I'm interested in nanotube
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what is system testing?
AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
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not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
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I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
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Prasenjit Reply
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.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Can someone give me problems that involes radical expressions like area,volume or motion of pendulum with solution
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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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