THE BIG 5 EQUATIONS OF MOTION

SPH3U February, 17 2015 at

$12 Speeding copCar Car

v= 30.0% a =3 .OM/s2

D= ? Vis Omlst . ? Vf = ?

d = ?

time =(t. 1)

* when the cop car catches

the speeding car they have

travelled the same distance'*

Speeding [ op CAR

r=ekI→d:Bad.ve/otttai'

D= yzatz>

CAR C 0 Pz

d = rt D= Yzat

¥Yz a * 2

¢o⇒t =

ke.oaspLtaj2Ct.ii.ct.Dtttxttttl.t2.2ttl@Atskl3.org

fleet ti )30 t = 1.562-2++1)

30 t = 1.5+2 . 3t t !

0 =/ .tt?3t.3ott 1.5

0 = 1.5+2-33++1,5Factor out 1.5

0 =L . S(t2 . 22T +1 )a =L b =

- 22 c= I

Quadratic Formula

to = - b±M÷Za

22+221 or 22-22

t.IE

How far ?

v. -

30

nlsid25

D= (301×225)D= 660 m

Worksheet ¥11

ftp.QHE?s:*-

= 9 .8ng[dad

Required 8i=2aoa[up]

Vt ,at

Analysis : Vj .vit2add

k=htadt

Feb . 23

Quiz tomorrow → By 5

+ Gravity

Pg .

56 ¥63+

--

9 [ up ]Adnkn

Given:

tftiszio!! Nsw

Retired Ed&

= 9845 [ down]

AnaYYiad5v.isttlg@XdtTbd.fKusXtkt9tHusjsd.s

21 m tt4.9¥. ) (4.41$)bd= -21 m

-

21.609NEd =- 42.609 m

Ed = - 43 m

• ! The bridge is 43 mabove

the river .

Pg .

56 ¥64

fEdtJp Rup ]~

Give Vi = 22 mls [ up ]vf = Omls ( at

yqyht)a

=

9.8 45 [down ]

Required : max. height ( Ed)

at

Analysis 2 :vi+2add

It.vt±Zabd=@Dtµ⇒÷

9.8ms ' )= -484n÷÷%bd=247m zd=25m

Solving for It

* max . height occurs halfway

through flight w multiplyat by 2

.

*

Eds ( Itsy at

staidqt

'YF÷It at ? 2.24 s

At = 2 ( 247 n) : . 4.48 s

- :.

4.5522 Ys TO m/s

Worksheet #9 → +

Given :FWD*

- 2.3 s

I= 25km/h [ FWD ]Jz .

. 62km/h [ FWD ]

Required Z manHIS

Analysis E. rT±.

Steps : a- = 62¥ .

25kg.

2.3 S

E. 37¥÷ s

E- 16 ¥:←"EI¥÷¥.

sa = 4.5 m/i

Prepare for QuizE.

Read pgs .

5875