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In addition to being useful in problem solving, the equation v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {} gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that

  • final velocity depends on how large the acceleration is and how long it lasts
  • if the acceleration is zero, then the final velocity equals the initial velocity ( v = v 0 ) size 12{ \( v=v rSub { size 8{0} } \) } {} , as expected (i.e., velocity is constant)
  • if a size 12{a} {} is negative, then the final velocity is less than the initial velocity

(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.)

Making connections: real-world connection


Space shuttle blasting off at night.
The Space Shuttle Endeavor blasts off from the Kennedy Space Center in February 2010. (credit: Matthew Simantov, Flickr)

An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified—short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.

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

We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with

v = v 0 + at . size 12{v=v rSub { size 8{0} } + ital "at"} {}

Adding v 0 size 12{v rSub { size 8{0} } } {} to each side of this equation and dividing by 2 gives

v 0 + v 2 = v 0 + 1 2 at . size 12{ { {v rSub { size 8{0} } +v} over {2} } =v rSub { size 8{0} } + { {1} over {2} } ital "at" "." } {}

Since v 0 + v 2 = v - size 12{ { {v rSub { size 8{0} } +v} over {2} } = { bar {v}}} {} for constant acceleration, then

v - = v 0 + 1 2 at . size 12{ { bar {v}}=v rSub { size 8{0} } + { {1} over {2} } ital "at" "." } {}

Now we substitute this expression for v - size 12{ { bar {v}}} {} into the equation for displacement, x = x 0 + v - t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {} , yielding

x = x 0 + v 0 t + 1 2 at 2 ( constant a ) . size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } " " \( "constant "a \) "." } {}

Calculating displacement of an accelerating object: dragsters

Dragsters can achieve average accelerations of 26 . 0 m/s 2 size 12{"26" "." "0 m/s" rSup { size 8{2} } } {} . Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

Dragster accelerating down a race track.
U.S. Army Top Fuel pilot Tony “The Sarge” Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.)

Strategy

Draw a sketch.

Acceleration vector arrow pointing toward the right in the positive x direction, labeled a equals twenty-six point 0 meters per second squared. x position graph with initial position at the left end of the graph. The right end of the graph is labeled x equals question mark.

We are asked to find displacement, which is x if we take x 0 size 12{x rSub { size 8{0} } } {} to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use 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} } } {} once we identify v 0 size 12{v rSub { size 8{0} } } {} , a size 12{a} {} , and t size 12{t} {} from the statement of the problem.

Solution

1. Identify the knowns. Starting from rest means that v 0 = 0 size 12{v rSub { size 8{0} } =0} {} , a size 12{a} {} is given as 26 . 0 m/s 2 size 12{"26" "." 0`"m/s" rSup { size 8{2} } } {} and t size 12{t} {} is given as 5.56 s.

2. Plug the known values into the equation to solve for the unknown x size 12{x} {} :

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} } "." } {}

Since the initial position and velocity are both zero, this simplifies to

x = 1 2 at 2 . size 12{x= { {1} over {2} } ital "at" rSup { size 8{2} } "." } {}

Substituting the identified values of a size 12{a} {} and t size 12{t} {} gives

x = 1 2 26 . 0 m/s 2 5 . 56 s 2 , size 12{x= { {1} over {2} } left ("26" "." "0 m/s" rSup { size 8{2} } right ) left (5 "." "56 s" right ) rSup { size 8{2} } ,} {}

yielding

x = 402 m. size 12{x="402 m"} {}

Discussion

If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.

Questions & Answers

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
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?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
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im not good at math so would this help me
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yes
Asali
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what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
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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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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
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what is system testing?
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
good afternoon madam
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
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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
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Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
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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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Source:  OpenStax, Phy 2053 - general physics i. OpenStax CNX. Jul 09, 2012 Download for free at http://cnx.org/content/col11435/1.1
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