CK-12-Physics-Concepts - Intermediate

http://www.ck12.org Chapter 2. Motion in a Straight Line

vave=^12 (vf+vi)

The distance, then, for uniformly accelerating motion can be found by multiplying the average velocity by the time.

d=^12 (vf+vi)(t) (Equation 1 )

We know that the final velocity for constantly accelerated motion can be found by multiplying the acceleration times
time and adding the result to the initial velocity,vf=vi+at.

The second equation that relates displacement, time, initial velocity, and final velocity is generated by substituting
this equation into equation 1.

Start by distributing the 1/2 in equation 1 through:

d=

1

2

(vf+vi)(t) =

1

2

vft+

1

2

vit

We know thatvf=vi+at. Therefore:
1
2 vit+

1
2 (t)(vi+at)
=^12 vit+^12 vit+^12 at^2

=vit+^12 at^2 (Equation 2)

The third equation is formed by combiningvf=vi+atandd=^12 (vf+vi)(t). If we solve the first equation for t
and then substitute into the second equation, we get

d=

( 1

2

)

(vf+vi)

(v

f−vi

a

)

=

( 1

2

)(v^2 f−v^2 i

a

)

And solving forvf^2 yields

vf^2 =vi^2 + 2 ad (Equation 3 )

Keep in mind that these three equations are only valid when acceleration is constant. In many cases, the initial
velocity can be set to zero and that simplifies the three equations considerably. When acceleration is constant and
the initial velocity is zero, the equations can be simplified to:

d=

1

2

vft

d=

1

2

at^2 and

vf^2 = 2 ad.

Example: Suppose a planner is designing an airport for small airplanes. Such planes must reach a speed of 56 m/s
before takeoff and can accelerate at 12.0 m/s^2. What is the minimum length for the runway of this airport?

Solution: The acceleration in this problem is constant and the initial velocity of the airplane is zero. Therefore,
we can use the equationvf^2 = 2 adand solve ford.

d=vf

2

2 a=

( 56 m/s)^2

( 2 )( 12. 0 m/s^2 )

=130 m

Example: How long does it take a car to travel 30.0 m if it accelerates from rest at a rate of 2.00 m/s^2?