Monday Oct 17 Physics

1) Try This Word Problem in Class:

Marvin the marble starts from rest with an acceleration of 3 m/s2. Sammy the sphere starts 2 m behind him with an acceleration of 5 m/s2. How far does each travel before they collide?

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

D=ViT + 1/2AT2         D= 0 + ½(3)T2                        D=1.5T2

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

D+2=ViT + 1/2AT2     D= 0 + ½(5)T2            -2         D=2.5T2-2

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

Sammy                        =                      Marv

2.5T2-2                        =                      1.5T2

1T2                              =          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

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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 + ½AT2   2=0+1/2A(3)2, 2=4.5A, A=.4444 m/s2

OR A=(1.333-0)/3=.444 m/s2

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 + ½AT2   2=0+1/2A(4)2, 2=8A, A=.25 m/s2

OR A=(1-0)/4=.25 m/s2

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/s2, Vi=0 m/s D=????

Eq 4 , D= ViT+1/2AT2     D= 0+ ½(.25)(3)2 = 1.125 m to make ball2 collide with ball1

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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/s2.

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+1/2AT2=20T +1/2(0)T2 = 30T = D

Police:             Vi=0, A=2.44,             D =ViT+1/2AT2= (0)T + ½ (2.44)T2 = 1.22T2

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

speeder            =          police

30 T                 =          1.22T2

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

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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/s2. 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+1/2AT2

60=20T+1/2(3)T2

0=1.5T2+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/s2. How far do I have to travel before I stop?

Vi=30 m/s, A = -2 m/s2, Vf=0  D=???

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+1/2AT2

35=50T+1/2(-2)T2

0=-1T2+50T-35 or 0= T2-50T+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/s2. 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+1/2AT2= (0)T + ½ (3)T2 = 1.5T2

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

Vf2=Vi2 + 2 A D = 2 (3)(4), Vf = 4.899 m/s at the end of the first part

8-4=4= D =ViT+1/2AT2= (0)T + ½ (3)T2 so 4=1.5T2, 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

1.5T2               =          4+4+ 4.899 (T-1.633)

0          =          1.5T2 – 4.899 T -4.899(-1.633)-8

0          =          1.5T2 – 4.899 T -4.899(-1.633)-8

0          =          1.5T2 – 4.899 T           +8-8

0          =          1.5T2 – 4.899 T           +0

T= 3.266 sec