Changing MotionMathematical Application

Constant Motion vs Changing Motion

change in velocity

∆ 𝒗

vivf

∆v

∆ 𝑣=𝑣 𝑓 −𝑣 𝑖

acceleration𝒂𝑎=

∆𝑣∆ 𝑡

∆ta

Rematch

Exampl

e1

A car driving east at 33 m/s accelerates to a velocity of 42 m/s in 6.0 seconds. What is the car’s acceleration?

G:

U:

E:

S:

S:

E:

S:

S:

vi vf

∆v

∆t

a

(draw picture)

𝒙 𝒊 𝒙 𝒇

∆𝒅

Combining Formulas

∆ 𝑣=𝑣 𝑓 −𝑣 𝑖 𝑎=∆𝑣∆ 𝑡

𝑎= ∆𝑡∆ 𝑣𝑣 𝑓 −𝑣 𝑖

Exampl

e 2A second car, also driving east, has a velocity of 20 m/s. It accelerates to a velocity of 50 m/s in 15 seconds. What is the second car’s acceleration?

G:

U:

E:

S:

S:

vi vf

∆v

∆t

a

(draw picture)

𝒙 𝒊 𝒙 𝒇

∆𝒅

Formulas = Relationships Between Variables

∆𝒅=𝒙 𝒇 −𝒙 𝒊∆𝒅𝒙 𝒊 𝒙 𝒇

𝒗 ∆ 𝒕∆𝒅 𝒗=∆𝒅∆ 𝒕

∆ 𝒗∆ 𝒕 𝒂 𝒂=∆𝒗∆ 𝒕

∆ 𝒗𝒗 𝒊𝒗 𝒇 ∆ 𝒗=𝒗 𝒇 −𝒗𝒊

𝒗 𝒇 𝒗𝒊∆ 𝒕 𝒂 𝒂=𝒗 𝒇 −𝒗𝒊

∆ 𝒕

Always Solve For Your Unknown Variable!

𝒗 𝒇=𝒗𝒊+𝒂∆ 𝒕

𝒗 𝒊=𝒗 𝒇 −𝒂∆ 𝒕

∆ 𝒕=𝒗 𝒇 −𝒗 𝒊

𝒂

𝒂=𝒗 𝒇 −𝒗𝒊

∆ 𝒕

Exampl

e 3A runner starts from rest and accelerates at 1.6 m/s2. How fast is he running after 4.8 seconds?

G:

U:E:

S:

S: +𝒗𝒊+𝒗𝒊

vi vf

∆v

∆t

a

(draw picture)

𝒙 𝒊 𝒙 𝒇

∆𝒅

What if You Have a Different Variable?

𝒂=(𝒗 𝒇

𝟐−𝒗𝒊𝟐 )

𝟐∆𝒅

𝒗 𝒊=√(𝒗 𝒇𝟐−𝟐𝒂∆𝒅 )

∆𝒅=(𝒗 𝒇

𝟐−𝒗𝒊𝟐 )

𝟐𝒂

𝒗 𝒇❑=√(𝒗 𝒊

𝟐+𝟐𝒂∆𝒅 )

∆ 𝒕

𝒗 𝒇𝟐=𝒗𝒊

𝟐+𝟐𝒂∆𝒅𝒗 𝒇

❑=𝒗𝒊❑+𝒂∆ 𝒕

Exampl

e 4A man driving his car at 54 m/s sees a deer in the road 250 meters away. He slams his brakes and accelerates at -6.4m/s2. Does the man stop before he hits the deer?

𝒗 𝒊 𝒗 𝒇a

(draw picture)

𝒙 𝒊 𝒙 𝒇

∆𝒅

∆ 𝒗

𝒙𝒅

𝒅𝒅

𝐈𝐅 :∆𝒅 ≥𝒅𝒅𝐓𝐇𝐄𝐍 :

Exampl

e 4G:

U:E:

S:

S:

∆𝒅=(𝒗 𝒇

𝟐−𝒗𝒊𝟐 )

𝟐𝒂

A man driving his car at 54 m/s sees a deer in the road 250 meters away. He slams his brakes and accelerates at -6.4m/s2. Does the man stop before he hits the deer?

𝟐𝒂 𝟐𝒂 ∆𝒅<𝒅𝒅 ∴

Summary

initial velocityfinal velocitychange in velocityacceleration

distanceinitial positionfinal positiondisplacementinitial timefinal timetime intervalvelocity

variables formulas𝒅𝒕=𝒅𝟏+𝒅𝟐+𝒅𝟑

∆𝒅=𝒙 𝒇 −𝒙 𝒊∆ 𝒕=𝒕 𝒇 − 𝒕𝒊𝒗=

∆𝒅∆ 𝒕

𝒂=∆𝒗∆ 𝒕

𝒗 𝒇𝟐=𝒗𝒊

𝟐+𝟐𝒂∆𝒅𝒗 𝒇

❑=𝒗𝒊❑+𝒂∆ 𝒕