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The equation v - = v 0 + v 2 size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } } {} reflects the fact that, when acceleration is constant, v size 12{v} {} is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation v - = v 0 + v 2 size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } } {} to check this, we see that

v - = v 0 + v 2 = 30 km/h + 60 km/h 2 = 45 km/h, size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } = { {"30 km/h"+"60 km/h"} over {2} } ="45 km/h,"} {}

which seems logical.

Calculating displacement: how far does the jogger run?

A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?

Strategy

Draw a sketch.

Velocity vector arrow labeled v equals 4 point zero zero meters per second over an x axis displaying initial and final positions. Final position is labeled x equals question mark.

The final position x size 12{x} {} is given by the equation

x = x 0 + v - t . size 12{x=x rSub { size 8{0} } + { bar {v}}t} {}

To find x size 12{x} {} , we identify the values of x 0 size 12{x rSub { size 8{0} } } {} , v - size 12{ { bar {v}}} {} , and t size 12{t} {} from the statement of the problem and substitute them into the equation.

Solution

1. Identify the knowns. v - = 4 . 00 m/s , Δ t = 2 . 00 min size 12{Δt=2 "." "00 min"} {} , and x 0 = 0 m size 12{x rSub { size 8{0} } ="0 m"} {} .

2. Enter the known values into the equation.

x = x 0 + v - t = 0 + 4 . 00 m/s 120 s = 480 m size 12{x=x rSub { size 8{0} } + { bar {v}}t=0+ left (4 "." "00 m/s" right ) left ("120 s" right )="480 m"} {}

Discussion

Velocity and final displacement are both positive, which means they are in the same direction.

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The equation x = x 0 + v - t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {} gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on v - size 12{ { bar {v}}} {} rather than on v - size 12{ { bar {v}}} {} raised to some other power, such as v - 2 size 12{ { bar {v}} rSup { size 8{2} } } {} . When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45 km/h.

Line graph showing displacement in meters versus average velocity in meters per second. The line is straight with a positive slope. Displacement x increases linearly with increase in average velocity v.
There is a linear relationship between displacement and average velocity. For a given time t size 12{t} {} , an object moving twice as fast as another object will move twice as far as the other object.

Solving for final velocity

We can derive another useful equation by manipulating the definition of acceleration.

a = Δ v Δ t

Substituting the simplified notation for Δ v and Δ t gives us

a = v v 0 t ( constant a ) . size 12{a= { {v - v rSub { size 8{0} } } over {t} } " " \( "constant "a \) "." } {}

Solving for v size 12{v} {} yields

v = v 0 + at ( constant a ) . size 12{v=v rSub { size 8{0} } + ital "at"" " \( "constant "a \) "." } {}

Calculating final velocity: an airplane slowing down after landing

An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1 . 50 m/s 2 size 12{1 "." "50 m/s" rSup { size 8{2} } } {} for 40.0 s. What is its final velocity?

Strategy

Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.

Velocity vector arrow pointing toward the right in the positive x direction. Initial velocity equals seventy meters per second. Final velocity equals question mark. An acceleration vector arrow pointing toward the left labeled a equals negative 1 point 50 meters per second squared.

Solution

1. Identify the knowns. v 0 = 70 . 0 m/s size 12{Δv="70" "." "0 m/s"} {} , a = 1 . 50 m/s 2 size 12{a= - 1 "." "50 m/s" rSup { size 8{2} } } {} , t = 40 . 0 s .

2. Identify the unknown. In this case, it is final velocity, v f size 12{v rSub { size 8{f} } } {} .

3. Determine which equation to use. We can calculate the final velocity using the equation v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {} .

4. Plug in the known values and solve.

v = v 0 + at = 70 . 0 m/s + 1 . 50 m/s 2 40 . 0 s = 10 . 0 m/s size 12{v=v rSub { size 8{0} } + ital "at"="70" "." "0 m/s"+ left ( - 1 "." "50 m/s" rSup { size 8{2} } right ) left ("40" "." "0 s" right )="10" "." "0 m/s"} {}

Discussion

The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.

An airplane moving toward the right at two points in time. At time equals 0 the velocity vector arrow points toward the right and is labeled seventy meters per second. The acceleration vector arrow points toward the left and is labeled negative 1 point 5 meters per second squared. At time equals forty seconds, the velocity arrow is shorter, points toward the right, and is labeled ten meters per second. The acceleration vector arrow is still pointing toward the left and is labeled a equals negative 1 point 5 meters per second squared.
The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Note that the acceleration is negative because its direction is opposite to its velocity, which is positive.
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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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