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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.

Got questions? Get instant answers now!

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

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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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