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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

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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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