SPH3U February, 17 2015 THE BIG 5 EQUATIONS OF MOTION
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THE BIG 5 EQUATIONS OF MOTION SPH3U February, 17 2015 at
Transcript of SPH3U February, 17 2015 THE BIG 5 EQUATIONS OF MOTION
THE BIG 5 EQUATIONS OF MOTION
SPH3U February, 17 2015 at
$12 Speeding copCar Car
v= 30.0% a =3 .OM/s2
D= ? Vis Omlst . ? Vf = ?
d = ?
time =(t. 1)
* when the cop car catches
the speeding car they have
travelled the same distance'*
Speeding [ op CAR
r=ekI→d:Bad.ve/otttai'
D= yzatz>
CAR C 0 Pz
d = rt D= Yzat
¥Yz a * 2
¢o⇒t =
fleet ti )30 t = 1.562-2++1)
30 t = 1.5+2 . 3t t !
0 =/ .tt?3t.3ott 1.5
0 = 1.5+2-33++1,5Factor out 1.5
0 =L . S(t2 . 22T +1 )a =L b =
- 22 c= I
Quadratic Formula
to = - b±M÷Za
22+221 or 22-22
t.IE
How far ?
v. -
30
nlsid25
D= (301×225)D= 660 m
Worksheet ¥11
ftp.QHE?s:*-
= 9 .8ng[dad
Required 8i=2aoa[up]
Vt ,at
Analysis : Vj .vit2add
k=htadt
Feb . 23
Quiz tomorrow → By 5
+ Gravity
Pg .
56 ¥63+
--
9 [ up ]Adnkn
Given:
tftiszio!! Nsw
Retired Ed&
= 9845 [ down]
21 m tt4.9¥. ) (4.41$)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
