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What else can we learn by examining the equation x = x 0 + v 0 t + 1 2 at 2 ? size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {} We see that:

  • displacement depends on the square of the elapsed time when acceleration is not zero. In [link] , the dragster covers only one fourth of the total distance in the first half of the elapsed time
  • if acceleration is zero, then the initial velocity equals average velocity ( v 0 = v - size 12{v rSub { size 8{0} } = { bar {v}}} {} ) and x = x 0 + v 0 t + 1 2 at 2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {} becomes x = x 0 + v 0 t size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t} {}

Solving for final velocity when velocity is not constant ( a 0 )

A fourth useful equation can be obtained from another algebraic manipulation of previous equations.

If we solve v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {} for t size 12{t} {} , we get

t = v v 0 a . size 12{t= { {v - v rSub { size 8{0} } } over {a} } "." } {}

Substituting this and v - = v 0 + v 2 size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } } {} into x = x 0 + v - t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {} , we get

v 2 = v 0 2 + 2 a x x 0 ( constant a ) . size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a left (x - x rSub { size 8{0} } right )" " \( "constant "a \) "." } {}

Calculating final velocity: dragsters

Calculate the final velocity of the dragster in [link] without using information about time.

Strategy

Draw a sketch.

Acceleration vector arrow pointing toward the right, labeled twenty-six point zero meters per second squared. Initial velocity equals 0. Final velocity equals question mark.

The equation v 2 = v 0 2 + 2 a ( x x 0 ) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.

Solution

1. Identify the known values. We know that v 0 = 0 size 12{v rSub { size 8{0} } =0} {} , since the dragster starts from rest. Then we note that x x 0 = 402 m size 12{x - x rSub { size 8{0} } ="402 m"} {} (this was the answer in [link] ). Finally, the average acceleration was given to be a = 26 . 0 m/s 2 size 12{a="26" "." "0 m/s" rSup { size 8{2} } } {} .

2. Plug the knowns into the equation v 2 = v 0 2 + 2 a ( x x 0 ) and solve for v .

v 2 = 0 + 2 26 . 0 m/s 2 402 m . size 12{v rSup { size 8{2} } =0+2 left ("26" "." "0 m/s" rSup { size 8{2} } right ) left ("402 m" right )} {}

Thus

v 2 = 2 . 09 × 10 4 m 2 /s 2 . size 12{v rSup { size 8{2} } =2 "." "09" times "10" rSup { size 8{4} } `m rSup { size 8{2} } "/s" rSup { size 8{2} } } {}

To get v size 12{v} {} , we take the square root:

v = 2 . 09 × 10 4 m 2 /s 2 = 145 m/s .

Discussion

145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.

An examination of the equation v 2 = v 0 2 + 2 a ( x x 0 ) size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a \( x - x rSub { size 8{0} } \) } {} can produce further insights into the general relationships among physical quantities:

  • The final velocity depends on how large the acceleration is and the distance over which it acts
  • For a fixed deceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)

Putting equations together

In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.

Summary of kinematic equations (constant a size 12{a} {} )

x = x 0 + v - t size 12{x=`x rSub { size 8{0} } `+` { bar {v}}t} {}
v - = v 0 + v 2 size 12{ { bar {v}}=` { {v rSub { size 8{0} } +v} over {2} } } {}
v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {}
x = x 0 + v 0 t + 1 2 at 2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}
v 2 = v 0 2 + 2 a x x 0 size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a left (x - x rSub { size 8{0} } right )} {}

Calculating displacement: how far does a car go when coming to a halt?

On dry concrete, a car can decelerate at a rate of 7 . 00 m/s 2 size 12{7 "." "00 m/s" rSup { size 8{2} } } {} , whereas on wet concrete it can decelerate at only 5 . 00 m/s 2 size 12{5 "." "00 m/s" rSup { size 8{2} } } {} . Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.

Strategy

Draw a sketch.

Initial velocity equals thirty meters per second. Final velocity equals 0. Acceleration dry equals negative 7 point zero zero meters per second squared. Acceleration wet equals negative 5 point zero zero meters per second squared.

In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.

Questions & Answers

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
Commplementary angles
Idrissa Reply
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Sherica
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Sherica
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Tamia
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Uday
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+_
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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.
Al
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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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J, combine like terms 7x-4y
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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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Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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AMJAD
preparation of nanomaterial
Victor Reply
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
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Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
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Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
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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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Source:  OpenStax, Physics 105: adventures in physics. OpenStax CNX. Dec 02, 2015 Download for free at http://legacy.cnx.org/content/col11916/1.1
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