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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.

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Questions & Answers

example ofchange of state of the body in the effectof heat
Abiodun Reply
what is normal force?
Neyaz Reply
the force that pushes upward on us. the force that opposes gravity
clifford
upthrust of air
Abdikadir
Newton's 3rd law. the force of the ground (earth) that pushes back on gravity, keeping us on the ground instead of sinking into it.
clifford
I really need lots of questions on frictional force
Ogboru Reply
Questions or answers?
Shii
I can help answering what I can
Shii
does friction also need some force to perform?
Mohit
no friction is a force just like the gravitational force
clifford
yeah but u can't apply friction anywhere else like other forces
Mohit
I don't understand that question. friction does work alongside other forces based on the situation.
clifford
eg. when walking there are two forces acting on us gravitational and frictional force. friction helps us move forward and gravity keeps us on the ground
clifford
friction is a contact force. Two surfaces are necessary for the force to work.
clifford
hope this helped
clifford
the friction force which oppose while it contact with surrounding. there are two kind of friction. slidding and rolling friction.
Neyaz
What is physics?
Jeuloriz Reply
physics is a branch of science in which we are dealing with the knowledge of our physical things. macroscopic as well as microscopic. we are going look inside the univers with the help of physics. you can learn nature with the help of physics. so many branches of physics you have to learn physics.
vijay
What are quarks?
Breanna Reply
6 type of quarks
Neyaz
what is candela
Akani Reply
Candela is the unit for the measurement of light intensity.
Osei
any one can prove that 1hrpower= 746 watt
Neyaz Reply
Newton second is the unit of ...............?
Neyaz
Impulse and momentum
Fauzia
force×time and mass× velocity
vijay
Good
Neyaz
What is the simple harmonic motion?
Fauzia Reply
oscillatory motion under a retarding force proportional to the amount of displacement from an equilibrium position
Yuri
Straight out of google, you could do that to, I suppose.
Yuri
*too
Yuri
ok
Fauzia
Oscillatory motion under a regarding force proportional to the amount of displacement from an equilibrium position
Neyaz
examples of work done by load of gravity
Maureen Reply
What is ehrenfest theorem?
Fauzia Reply
You can look it up, faster and more reliable answer.
Yuri
That isn't a question to ask on a forum and I also have no idea what that is.
Yuri
what is the work done by gravity on the load 87kj,11.684m,mass xkg[g=19m/s
Maureen
What is law of mass action?
Fauzia Reply
rate of chemical reactions is proportional to concentration of reactants ...
muhammad
ok thanks
Fauzia
what is lenses
Ndobe Reply
lenses are two types
Fauzia
concave and convex
muhammad
right
Fauzia
speed of light in space
Vikash Reply
in vacuum speed of light is 3×10^8 m/s
vijay
ok
Vikash
2.99×10^8m/s
Umair
2.8820^8m/s
Muhammed
which is correct answer
Vikash
he is correct but we can round up in simple terms
vijay
3×10^8m/s
vijay
is it correct
Fauzia
I mean 3*10^8 m/s ok
vijay
299792458 meter per second
babar
3*10^8m/s
Neyaz
how many Maxwell relations in thermodynamics
vijay
how we can do prove them?
vijay
What is second law of thermodynamics?
Neyaz
please who has a detailed solution to the first two professional application questions under conservation of momentum
Kwaku Reply
I want to know more about pressure
Osei
I can help
Emeh
okay go on
True
I mean on pressure
Emeh
definition of Pressure
John
it is the force per unit area of a substance.S.I unit is Pascal 1pascal is defined as 1N acting on 1m² area i.e 1pa=1N/m²
Emeh
pls explain Doppler effect
Emmex
solve this an inverted differential manometer containing oil specific gravity 0.9 and manometer reading is 400mm find the difference of pressure
Abayomi Reply

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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