Vertical Motion

Circular Motion

Vector’s Components Velocity has both the magnitude and direction (is a vector).

For motion we need to know: distance, time, speed, velocity, acceleration.

2

The average speed is the total distance divided by the total time or change in

distance divided by change in time:

Instantaneous speed is change in distance divided by very short time

(t ! zero):

Both average and instantaneous speed have only a magnitude (is a scalar).

totaltotalavtdv /=

0/ !"""= ttdvinst

tdvav

!!= /

The acceleration shows how fast the object changes its velocity;

the change in velocity divided by the change in time: tva !!= /

The distance traveled by an object at constant acceleration: d = v0t +

at2

2

3

Types of Motion

1. Horizontal motion

2. Vertical motion

3. Circular motion

(a) Motion at constant speed

(b) Motion at constant acceleration

constv =

consta =

4. Projectile motion

v!

v!

v!

v!

v!

v!

A car traveled 40 m for 4 seconds at constant acceleration from the rest.

Find the acceleration of the car.

Distance and Acceleration

1) 2 m/s2

2) 4 m/s2

3) 5 m/s2

4) 10 m/s2

4

2

22/5

16

80

)4(

)40)(2(sm

s

m

s

m===

2

2at

d =

2

2

t

da =

" An extremely important case of constant acceleration motion.

Vertical Motion, Free Fall

d =gt

2

2

" This means that any two objects, regardless of the mass or material

of the objects, released from a given height will take the same timeto reach the floor.

" When an object is released in free fall, it falls down with a constant

acceleration a = g = 9.8 m/s2 which is always directed down! (NB weare assuming here that air resistance is negligible)

Distance for free fall:

v = gtVelocity for free fall:

5

Falling Object

If you dropdrop a ball, it will have:

– zero initial velocity;

– the acceleration of 9.8 m/s2

due to gravity.

+

289 m/s.ga ==

6

The velocity and position can be

written as:

2

2gtd =gtv =

The velocity and acceleration have

the same direction (downward).

Throwing Object Upward

If you throwthrow a ball upward, it will have:

– a non-zero initial velocity;

– the acceleration of 9.8 m/s2 due to

gravity.

7

+

gtvv !=0

2

2

0

gttvd !=

At the highest point of elevation, the speed

is zero while the acceleration is 9.8 m/s2 !

The velocity and acceleration have the opposite

direction, upward and downward, respectively.

The ball is dropped from the 40 m high building.

Falling Ball

1. Find its speed after 2 seconds:

2. How long it takes time to reach the ground?

3. Find its speed near the ground?

8

smssmgtv /6.19)2)(/8.9( 2===

2

2gtd =

22 gtd = s

sm

m

g

dt 86.2

/8.9

)40)(2(22===

smssmgtv /28)86.2)(/8.9( 2===

A ball is thrown upward with the initial speed of 25 m/s.

Ball Throwing Upward

2. Find speed of the ball after 2 seconds:

1. How long it takes time to reach the highest point?

3. Find the distance from the ground to the highest point:

9

gtvv !=0

mssm

ssm 89.312

)55.2)(/8.9()55.2)(/25(

22

=!=

0== hpvv ssm

sm

g

vt 55.2

/8.9

/25

2

0===

gtvv !=0

2

2gttvd o !=

smssmsm /4.5)2)(/8.9(/25 2=!=

gtv =0

Circular Motion: Centripetal (Radial) Acceleration

This type of acceleration occurs when an

object moving on a circular path.v

v

v

v

rada

rada

rada

10

rada

rada

rada

rada

v

v

v

v

Acceleration is directed to the center, so-

called “centripetal acceleration” (the word

centripetal is derived from two Greek words

“seeking the center”) or “radial acceleration”.

The magnitude of the centripetal (radial)

acceleration: r

vaa radlcentripita

2

==

courve theof radius speed, == rv

r

r

v

rada

Circular Motion: Period and Frequency

fTor

Tf

11==

11

In a circular motion the time required for one rotation is the period, T

The unit of period: second (s)

For a circular motion or other periodic process we also can use the frequency, f

A wheel makes one revolution during 0.02 s. Find the frequency of its

rotation.

HzssT

f 501

5002.0

11====

The unit of frequency: 1/second = hertz ( Hz)

Centripetal (Radial) Acceleration

A ball rotates on a circle with a radius of 2 m at a speed of 4 m/s. Its

centripetal (radial) acceleration is:

1) 2 m/s

2) 2 m/s2

3) 8 m/s

4) 8 m/s2

22

/82

)/4(sm

m

sm==

r

vacentr

2

=

)radius(r

)velocity(v!

)onaccelerati

radialor lcentripeta(a!