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v = v 0 + v 2 = 40 km/h + 80 km/h 2 = 60 km/h .

In part (b), acceleration is not constant. During the 1-h interval, velocity is closer to 80 km/h than 40 km/h. Thus, the average velocity is greater than in part (a).

Graph A shows velocity in kilometers per hour plotted versus time in hour. Velocity increases linearly from 40 kilometers per hour at 1 hour, point vo, to 80 kilometers per hour at 2 hours, point v. Graph B shows velocity in kilometers per hour plotted versus time in hour. Velocity increases from 40 kilometers per hour at 1 hour, point vo, to 80 kilometers per hour at 2 hours, point v. Increase is not linear – first velocity increases very fast, then increase slows down.
(a) Velocity-versus-time graph with constant acceleration showing the initial and final velocities v 0 and v . The average velocity is 1 2 ( v 0 + v ) = 60 km / h . (b) Velocity-versus-time graph with an acceleration that changes with time. The average velocity is not given by 1 2 ( v 0 + v ) , but is greater than 60 km/h.

Solving for final velocity from acceleration and time

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 ) .

Solving for v yields

v = v 0 + a t ( constant a ) .

Calculating final velocity

An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s 2 for 40.0 s. What is its final velocity?

Strategy

First, we identify the knowns: v 0 = 70 m/s, a = −1.50 m/s 2 , t = 40 s .

Second, we identify the unknown; in this case, it is final velocity v f .

Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknown in terms of the knowns. We calculate the final velocity using [link] , v = v 0 + a t .

Solution

Substitute the known values and solve:

v = v 0 + a t = 70.0 m/s + ( −1.50 m/ s 2 ) ( 40.0 s ) = 10.0 m/s.

[link] is a sketch that shows the acceleration and velocity vectors.

Figure shows airplane at two different time periods. At t equal zero seconds it has velocity of 70 meters per second and acceleration of -1.5 meters per second squared. At t equal 40 seconds it has velocity of 10 meters per second and acceleration of -1.5 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 the acceleration is negative because its direction is opposite to its velocity, which is positive.

Significance

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

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In addition to being useful in problem solving, the equation v = v 0 + a t gives us insight into the relationships among velocity, acceleration, and time. 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 ), as expected (in other words, velocity is constant)
  • If a is negative, then the final velocity is less than the initial velocity

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

Solving for final position with constant acceleration

We can combine the previous equations 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 + a t .

Adding v 0 to each side of this equation and dividing by 2 gives

v 0 + v 2 = v 0 + 1 2 a t .

Since v 0 + v 2 = v for constant acceleration, we have

v = v 0 + 1 2 a t .

Now we substitute this expression for v into the equation for displacement, x = x 0 + v t , yielding

Practice Key Terms 1

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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