Amber Truhanovitch                                                                                                      September 24, 2007

Physics 2 week project                                                                                                  Group C

Problem: Tom throws a stone straight-up into the air. The stone falls down and hits Freddie the fish that swam at a constant rate toward the bridge as the rock was being thrown. We must find the speed of Freddie on his trip before he went to “sleep with the fishes”.

-Givens-

a-b         ΔXab=31.75m (ΔXbd-dc=ab):(218ft=66.45m-34.7m=31.75m)                           

Viab=0m/s           Vfab=0m/s           aab=-9.81m/s^2                               

b-c          ΔXbc=31.75 m    Vibc= 0m/s          abc=-9.81m/s^2

c-d          ΔXcd=34.7m        Vfcd=0m/s

b-d         ΔXbd= 281ft (12in/1ft)(2.54cm/1in)(1m/100cm)= 66.45m              

abd=-9.81m/s^2                                Vibd=0m/s

e-d         ΔXed=10.1m       Vfed=0m/s          

Known-     tab+tcd=ted

Strategy: First we must find out all of the times because we know that the time from a-b plus c-d is equal to time of e-d because in the problem it states they took place at the same time and ended the same. We also know that the time of s-b is equal to the time of b-c which makes it a lot easier on us to find the total time. Once we have the time from both we find the total time and then use it in equation 1 to find the speed of Freddie.

 

1.       Eq. 3: ΔX=Vi(t)+1/2(a)(t^2)

ΔXbd=Vibd (tbd)+1/2(abd)(tbd^2)

-66.45=0t+1/2 (-9.81)(tbd^2)

-66.45=-4.905 (tbd^2)

13.547= (tbd^2)

3.68s=tbd

                2.    Δtcd=Δtab

Eq. 3: ΔX=Vi(t)+1/2(a)(t^2)

ΔXbc=Vibc (tbc)+1/2(abc)(tbc^2)

-31.75=0t+1/2(-9.81) (tbc^2)

-31.75= -4.905 (tbc^2)

6.4729= (tbc^2)

2.54s = Δtcd=Δtab

                3.    ttot= 3.68+2.54 = 6.22s

 

                4.   Eq. 1: ΔX=(Vi+Vf/2)(t)

                     ΔXed=(Vied+Vfed/2)(ted)

                      10.1= ((Vied + 0)/2)(6.22)

                      1.62= (Vied/2)

                      3.25= Vied (CONSTANT)

                      

Speed of Freddie = 3.25 m/s

 
 
 

 


Amber Truhanovitch                                                                                                      September 28, 2008- Group C

Problem: Huey the hamster has been sent up in a model rocket. It is launched at rest and we must figure out how long it took Huey to complete the trip. We know that the engine burns for a specific amount of time with a constant acceleration.  Once at the top the rocket will free fall down until the parachute opens and at a constant speed travel to the ground.

-Givens-

a-b         tab= 2.11s                Viab= 0 m/s

Vfab= 125.7 mph = (5280ft/1mi)(12in/1ft)(2.54cm/1in)(1m/100cm)(1h/60min)(1m/60s)= 56.19m/s

b-c          Vibc= Vfab= 56.19 m/s                            Vfbc= 0 m/s             abc= -9.81 m/s ̂2      tbc= ?

c-d          vicd=0 m/s             Vicd=0 m/s             a=-9.81 m/s^2       

ΔX=476ft=(12in/1ft)(2.54cm/1in)(1m/100cm)= 145.0847m              tcd= ?s                   Vfcd= ?m/s

d-e         Vide=Vfcd= ? (-53.34m/s (step 4))    Vfde=-2.1m/s         ΔXde= Xab+Xbc =(75.115m (step 7))   a=0m/s^2     tde=?

Strategy:  I must find the total time for the trip, so I will start by solving all of the unknowns to find the time of each interval (i.e. a-b). In a-b we already have the time so I started on b-c. In b-c we have everything to find the time so I used eq.2 to find the time. From c-d we also had everything we needed to find the time because from b-c there is acceleration because the engine is no longer burning. After, I go ahead to find the Vf of c-d because it is the Vi of d-e. Now I’m on d-e I must first find the ΔX so that I can find the time. ΔX is equal to ΔXac- ΔXcd, once I have found the height I can go ahead and find the time. Now that we have all of the times and now we just need to add them all up to get the total time.

1.        tab= 2.11 (given)

2.        http://tbn0.google.com/images?q=tbn:LNm8XbxLyRHkzM:http://www1.istockphoto.com/file_thumbview_approve/2409200/2/istockphoto_2409200_sun_with_clouds_animation_cartoon.jpgEq.2: Vf=Vi+at

Vfbc=Vibc+abc(tbc)

 0=56.19+(-9.81)(t)

-56.19=-9.81(t)

http://tbn0.google.com/images?q=tbn:qlc4kIuGCOQSpM:http://www.lillyarts.com/assets/images/Friskylove1.gif5.7278s=tbc

       3.    Eq.3: ΔX=Vi+1/2 (a)(t^2)

                ΔXcd=Vicd+1/2 (acd)(tcd^2)

                -145.0884=0t+1/2(-9.81)(t^2)

                -145.0848=-4.905(t^2)

                29.578= t^2

                5.44s=t

       4.   Eq1: ΔX=(Vi+Vf/2)(t)

                ΔXcd=(Vicd+Vfcd/2)(tcd)

                -145.0848=(0+Vf/2)(5.44)

                -26.67=(vf/2)

                Vf= -53.34 m/s

       5.    Eq.1: ΔX=(Vi+Vf/2)(t)

                ΔXab=(Viab+Vfab/2)(tab)

                ΔX= (0+56.19/2)2.11

                ΔX=59.28m

       6.    Eq4: Vf^2=Vi^2+2aΔX

                Vfbc^2=Vibc^2+2abcΔXbc

                        0=56.19^2+2(-9.81) ΔXbc

                        -3157.3162=-19.62ΔXbc

                        ΔXbc=160.92m

         7.   ΔXac=160.92+59.28=220.20=ΔXce

                        ΔXde= ΔXce- ΔXdc

              220.20-145.0848= 75.115m

         8.  Eq.1: ΔX=(Vi+Vf/2)(t)

                ΔXde=(Vide+Vfde/2)(tde) =-75.115=((-53.34+-2.1)/2)(t) = -75.115=-27.72t = t=2.71s

Ttot= 15.99s
 
        9. Total Time = 2.11+5.7278+5.44+2.71 =