Monday Oct 17 Physics

1) Try This Word Problem in Class:

Marvin the marble starts from
rest with an acceleration of 3 m/s^{2}. Sammy the sphere starts 2 m
behind him with an acceleration of 5 m/s^{2}. How far does each travel
before they collide?

*Marvin Vi=0, A = 3, D=??, T=?? *

*D=ViT + 1/2AT ^{2} D=
0 + ½(3)T^{2} D=1.5T^{2}*

* *

*Sammy Vi=0, A= 5, Dsam=??=Dmarv+2, T=??*

*D+2=ViT + 1/2AT ^{2} D= 0 + ½(5)T^{2} -2 D=2.5T^{2}-2*

*Sammy hits Marv in the same place (disp) at the same
time so:*

* Sammy = Marv*

*2.5T ^{2}-2 = 1.5T^{2}*

^{ }1T^{2} = 2

* T = 1.414
sec*

* D=
1.5 (1.414) ^{2} = 3 meters (marv), Dsam=5 meters*

2) Go over any questions from pg 55, 58 (check answers)

3) Finish questions on and hand in Ramp Lab…

4) Wed/Thur Start Balls Collide (as part of Ramp Lab) so read ahead!

5) Homework, Word Problems on back (do on separate paper please)

6) Get yellow light project tomorrow/Thur

-------------------------------------------

Ball1 rolls down a 2m ramp in 3 seconds.

What is its initial
velocity? *Vi=0 m/s*

What is its average velocity?
(Eq 1 or 2) *Vavg=D/T=2/3=.667 m/s*

What is its final velocity?
(Eq 1 or 2) * Vavg=(vi+vf)/2
.667 = (0+vf)/2, Vf=1.333 m/s*

* *

What is its acceleration? (Eq
3 or 4) * D=ViT + ½AT ^{2} 2=0+1/2A(3)^{2}, 2=4.5A,
A=.4444 m/s^{2}*

*OR A=(1.333-0)/3=.444
m/s ^{2}*

* *

Ball2 rolls down a shallower
2m ramp in 4 seconds.

What is its initial
velocity? *Vi=0 m/s*

What is its average velocity?
(Eq 1 or 2) *Vavg=D/T=2/4=.5 m/s*

What is its final velocity?
(Eq 1 or 2) * Vavg=(vi+vf)/2
.5 = (0+vf)/2, Vf=1 m/s*

* *

What is its acceleration? (Eq
3 or 4) * D=ViT + ½AT ^{2} 2=0+1/2A(4)^{2}, 2=8A,
A=.25 m/s^{2}*

*OR A=(1-0)/4=.25 m/s ^{2}*

If I want Ball2 to collide
into Ball1, what will be the same between them? * TIME*

Use the TIME of Ball1 and the
ACCELERATION of Ball2 and find the Distance I have to place Ball2 at to make it
crash into Ball2.*Time = 3 seconds, A=.25 m/s ^{2}, Vi=0 m/s
D=???? *

*Eq 4 , D= ViT+1/2AT ^{2} D= 0+ ½(.25)(3)^{2}
= 1.125 m to make ball2 collide with ball1*

---------------------------------------------------

A speeder passes a parked police car at a constant 30 m/s.
The police car starts from rest with a uniform acceleration of 2.44 m/s^{2}.

a) How much time passes before the speeder is overtaken by the police car?

*Speeder: Vi=30,
Vf=30, D=VavgT D=
30T *

*or D
=ViT+1/2AT ^{2}=20T +1/2(0)T^{2} = 30T = D*

*Police: Vi=0,
A=2.44, D
=ViT+1/2AT ^{2}= (0)T + ½ (2.44)T^{2} = 1.22T^{2}*

* *

* “overtaken”
means same place (D) at same time (T) so*

* speeder = police*

* 30
T =
1.22T ^{2}*

* *

* one
solution is T=0, but we know they are in the same place then!*

* 30 = 1.22
T*

* T= 30/1.22 =
24.59 sec*

b) How far does the speeder get before being overtaken by the police car?

*D=30T
= 20 (24.59) = 737.7 meters*

------------------

I am 50 meters away from an intersection that is 10 m across
going 20 m/s. I see the light turn yellow. My car can accelerate at the rate of
3 m/s^{2}. What is the shortest time I can make it across the
intersection?

* Vi = 20
m/s, D=50+10=60 meters, A = 3*

*D =ViT+1/2AT ^{2}*

*60=20T+1/2(3)T ^{2}*

*0=1.5T ^{2}+20T-60*

*Using quadratic equation or math solver….*

*T= 2.522 sec or T=-15.86 so T=2.522 sec…*

4) I start off going 50 m/s and deaccelerate at the rate of
-2 m/s^{2}. How far do I have to travel before I stop?

*Vi=30 m/s, A = -2 m/s ^{2}, Vf=0 D=???*

*Vf ^{2}=Vi^{2}+2AD *

*0=(30) ^{2} +2(-2)D*

*-900= -4 D*

*D= 225 m*

**** (H)When will I be 35 m from the start?

*Vi = 50 m/s, D=35 meters, A = -2*

*D =ViT+1/2AT ^{2}*

*35=50T+1/2(-2)T ^{2}*

*0=-1T ^{2}+50T-35 or 0= T^{2}-50T+35*

*Using quadratic equation or math solver….*

*T= 49.28 sec or T= .71 both are correct (once going
and once in the reverse direction!)*

****(H) One car (Frank) starts 4 meters ahead of another
(Bob) . They both accelerate at the rate of 3 m/s^{2}. If car one
(Frank) stops accelerating after 4 meters, when will car two (Bob) catch up to
him?

*Bob: Di=0, Vi=0, A=3 D
=ViT+1/2AT ^{2}= (0)T + ½ (3)T^{2} = 1.5T^{2}*

* *

*Frank Di=4, Vi=0, A=3, Df =8 for the first part of the
journey*

* Vf ^{2}=Vi^{2}
+ 2 A D = 2 (3)(4), Vf = 4.899 m/s at the end of the first part*

* 8-4=4=
D =ViT+1/2AT ^{2}= (0)T + ½ (3)T^{2} so 4=1.5T^{2}, T
=1.633 for the first part to go 4 meters (8 meters from the start)*

* for
the second part of the journey, Frank went 4.899 m/s for (T-1.633) seconds, so
D= Vavg T = 8.9 (T-1.633) for the second part.*

* *

* So
Frank’s total displacement (after he stops accelerating) is *

*D= 4 (start)+4 (first part)+
4.899 (T-1.633)(even speed)*

* Bob
“catch up” to Frank in same place at same time:*

* Bob = Frank*

* 1.5T ^{2} = 4+4+
4.899 (T-1.633)*

* 0 = 1.5T ^{2}
– 4.899 T -4.899(-1.633)-8*

* 0 = 1.5T ^{2}
– 4.899 T -4.899(-1.633)-8*

*0 = 1.5T ^{2}
– 4.899 T +8-8*

* 0 = 1.5T ^{2}
– 4.899 T +0*

* T=
3.266 sec *