Motion & ForcesMotion & Forces

Part II. Describing Motion Motion Speed & Velocity Acceleration

C. AccelerationC. Acceleration

Acceleration the rate of change of velocitychange in speed or direction

t

vva if

a: acceleration

vf: final velocity

vi: initial velocity

t: time

a

vf - vi

t

C. AccelerationC. Acceleration

Positive acceleration “speeding up”

Negative acceleration (deceleration) “slowing down”

D. CalculationsD. CalculationsA roller coaster starts down a hill at 10 m/s.

Three seconds later, its speed is 32 m/s. What is the roller coaster’s acceleration?

GIVEN:

vi = 10 m/s

vf = 32 m/s

t = 3 s

a = ?

WORK:

a = (vf - vi) ÷ t

a = (32m/s - 10m/s) ÷ (3s)

a = 22 m/s ÷ 3 s

a = 7.3 m/s2a

vf - vi

t

D. CalculationsD. CalculationsA car is at a stop sign. It starts moving,

and in 3 minutes is going 45 miles per hour. What is the car’s acceleration?

GIVEN:

vi = 0 mph

vf = 45 mph

t = 3 mins = 0.05hrs

a = ?

WORK:

a = (vf - vi) ÷ t

a = (45 mph – 0mph) ÷ (.05 hours)

a = 45 mph ÷ .05 hrs.

a = 900 mpha

vf - vi

t

D. CalculationsD. CalculationsA cyclist bikes up a hill at an initial velocity

of 25 m/s. 30 seconds later, their velocity is 10 m/s. What is the cyclist’s acceleration?GIVEN:

vi = 25 m/s

vf = 10 m/s

t = 30 s

a = ?

WORK:

a = (vf - vi) ÷ t

a = (10m/s - 25m/s)÷(30s)

a = -15 m/s ÷ 30 s

a = -0.5 m/s2a

vf - vi

t

D. CalculationsD. CalculationsHow long will it take a car traveling 30 m/s

to come to a stop if its acceleration is -3 m/s2?

GIVEN:

vi = 30 m/s

vf = 0 m/s

a = -3 m/s2

t = ?

WORK:

t = (vf - vi) ÷ a

t = (0m/s-30m/s)÷(-3m/s2)

t = -30 m/s ÷ -3m/s2

t = 10 sa

vf - vi

t

0

100

200

300

400

0 5 10 15 20

Time (s)

Dis

tan

ce (

m)

Distance-Time Graph

E. Graphing MotionE. Graphing Motion

Acceleration is indicated by a curve on a Distance-Time graph.

Changing slope = changing velocity

E. Graphing MotionE. Graphing Motion

0

1

2

3

0 5 10

Sp

ee

d (m

/s)

Time (s)

Velocity-Time Graph

slope = acceleration + velocity = acceleration - velocity = deceleration (negative

acceleration)straight line (vertical) =

constant acceleration flat (horizontal) line =

no acceleration (constant velocity)

Ve

loc

ity

(m

/s)

C. Centripetal AccelerationC. Centripetal Acceleration

is the acceleration of objects traveling in a a circular motion

since the objects are always changing direction, so is the velocity

if the velocity is always changing, the object is accelerating (acceleration is a change in speed or direction)