Constant Acceleration

KinematicsMotion Equations

1

Constant Acceleration

Constant Acceleration Problem Solving

Equations of Motion

Centripetal and Tangential Acceleration

Free-Fall Motion


2

Motion can be determined by using a few simple equations.

rv

t

0fr r v t

The relationships between velocity and position are…

va

t

0fv v a t

The relationships between acceleration and velocity are…


3

If the acceleration is constant, we see that these become.

v a t 2

0

1

2r v t a t

Combining these, we find another equation.

tavv f

0

20 0 02f fv v v v v a t a a t

20 0 0 0 0

12 2

2f fv v v v a v t a t v v a r

0 0 2f fv v v v a r


4

We can use these three equations to solve for any motion involving constant acceleration.

v a t

20

1

2r v t a t

0 0 2f fv v v v a r

This equation relates velocity and time.

This equation relates position and time.

This equation relates position and velocity.


5

If we are only dealing with one vector component, then the equations become simple.

v a t

20

1

2r v t a t

0 0 2f fv v v v a r

Let’s just look at the x-component

x xv a t

20

1

2x xx v t a t

2 20 2xf x xv v a x

The y-component and z-component equations are similar.


6

Now let’s see how we use them.

Example: Two race cars are moving on a racetrack. The lead car is ahead by 10 m. Both cars are currently moving at 100 km/hr. If the second car accelerates at 10 m/s2, how long will it take to reach the lead car?

1r

2r

01 10mr

10

1

100 km/hr

0

v

a

20

2 2

100 km/hr

m 10

s

v

a

2

2 2

m 10

s

fv

a click the icon toopen the worksheet

Microsoft Excel Worksheet

1

1

100 km/hr

0

fv

a

2

http://web.nmsu.edu/~mdeanton/Kinematics/Acceleration2/Lecture%20Example.xls


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1r

2r

01 10mr

1 1 1x xv a t 2

1 1 0 1 1 1

1

2x xx v t a t 2 2

1 1 0 1 12xf x xv v a x

2 2 2x xv a t 2

2 2 0 2 2 2

1

2x xx v t a t 2 2

2 2 0 2 22xf x xv v a x

10

1

100 km/hr

0

v

a

20

2 2

100 km/hr

m 10

s

v

a

2


8


2 2 2x xv a t 2

2 2 0 2 2 2

1

2x xx v t a t

1 1 1x xv a t 2

1 1 0 1 1 1

1

2x xx v t a t 2 2

1 1 0 1 12xf x xv v a x

2 22

km m100 10

hr sxfv t

22 2 22

km 1 m100 10

hr 2 sx t t

22

2 22

km m100 2 10

hr sxfv x

1

km100 0

hrxfv

1 1

km100

hrx t

22

1

km100

hrxfv

2

10

1

100 km/hr

0

v

a

20

2 2

100 km/hr

m 10

s

v

a

1r

2r

01 10mr

2 22 2 0 2 22xf x xv v a x


9


2 22

km m100 10

hr sxfv t

22 2 22

km 1 m100 10

hr 2 sx t t

22

2 22

km m100 2 10

hr sxfv x

1

km100 0

hrxfv

1 1

km100

hrx t

22

1

km100

hrxfv

ttt 21

2 1 01x x x

2 2

km m100 10

hr sxfv t

22 2

km 1 m100 10

hr 2 sx t t

22

2 22

km m100 2 10

hr sxfv x

1

km100 0

hrxfv

2

km10 m 100

hrx t

22

1

km100

hrxfv

2

10

1

100 km/hr

0

v

a

20

2 2

100 km/hr

m 10

s

v

a

1r

2r

01 10mr


10


ttt 21

1 2

km m100 10

hr sxfv t

21 2

km 1 m100 10

hr 2 sx t t

22

1 12

km m100 2 10

hr sxfv x

2

km100 0

hrxfv

1

km10 m 100

hrx t

22

2

km100

hrxfv

2

10

1

100 km/hr

0

v

a

20

2 2

100 km/hr

m 10

s

v

a

1r

2r

01 10mr

2 1 01x x x


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ttt 21

21 2

km 1 m100 10

hr 2 sx t t

1

km10 m 100

hrx t

22

km km 1 m10 m 100 100 10

hr hr 2 st t t

2

10

1

100 km/hr

0

v

a

20

2 2

100 km/hr

m 10

s

v

a

1r

2r

01 10mr

2 1 01x x x


12


ttt 21

22

km km 1 m10 m 100 100 10

hr hr 2 st t t

22

1 m10 m 10

2 st

2

2

10 m2

m10

s

t

1.44 st

2

10

1

100 km/hr

0

v

a

20

2 2

100 km/hr

m 10

s

v

a

1r

2r

01 10mr

2 1 01x x x


13

What Happened to Centrifugal Force?

There is no such thing as centrifugal force.

So where did it come from?

A mistaken assumption is made that the forces on particles moving in a circle with constant speed have no forces acting on them.

Why is this false?

Acceleration comes from changes in velocity (direction, not just speed).

Circular motion requires acceleration and thus requires force.


14

centifugalF

T


The myth starts from the mistaken idea that there is no acceleration and therefore the total force is zero.

0centifugal amFT


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The fact is that velocity is changing and the acceleration is the centripetal acceleration. The force is NOT zero!!!!

lcentripetaamT

T


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Centripetal vs. Tangential Acceleration

Centripetal acceleration causes a particle to change its direction.

2

c

va

r

It points toward the center of the circle

r

v


17

Free Fall

Any particle, subject only to the force of gravity is in free-fall.

If an object is in free-fall and we define the positive y-axis as upward, then its acceleration is always given by

where g is the acceleration due to gravity and has a value of 9.81 m/s2 near the surface of the earth.

Note that the acceleration parallel to the earth’s surface is zero.

j ga


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Particle’s in free fall are subject only to the force of gravity.

Every particle in free-fall has an acceleration of 9.81 m/s2 downward.

Free Fall

The motion diagram for any object in free-fall that starts from rest is the same.


19

For particle’s that do not start at rest…

1.The vertical acceleration is 9.81 m/s2 downward.2.The horizontal acceleration is zero. (The horizontal velocity is constant.)

In other words, objects move in a very predictable way.But then, you already know this.

Free Fall


20

The motion of a baseball hit at an angle, undergoing free fall is a parabola.

Free Fall


21

The motion of a rocks thrown from a cliff at different horizontal speeds has some similarities.

Free Fall


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Equations

In free fall, the vertical position, velocity and acceleration are related by the equations

the horizontal position, velocity and acceleration are related by the equation

Note that we can write the components of the initial velocity as

20 2

1gttvy y gtvy

tvx x0 0 xv

000 cosvv x 000 sinvv y

Constant Acceleration

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