Caiti+C

__ **Week 2 (Jan 30th 2012 )** __


 * 01. A certain cab has a total run of 190 m and a maximum speed of 305 m/min. It accelerates from rest and back to rest at 1.22 m/s². **
 * i). How far does the cab move while accelerating to the maximum speed starting from rest? **
 * use the equation v^2=(v0)^2+2a(x-x0)
 * convert to SI units:
 * v=305 m/min*min/60 s= 5.08 m/s
 * substitute values:
 * (5.08 m/s)^2=(0 m/s)^2+2(1.22 m/s^2)(x-x0), then evaluate.
 * 25.81 m^2/s^2= (2.44 m/s^2)(x-x0)
 * divide both sides by (2.44 m/s^2).
 * x-x0=__distance while accelerating to maximum speed= 10.58 m.__
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? **
 * use equation v=v0+at.
 * let t1=time to reach 5.08 m/s and t2=time to return to rest, so t=time starting and ending at rest.
 * t1: substitute values:
 * 5.08 m/s=(1.22 m/s^2)*t
 * divide both sides by 1.22 m/s^2, so t1=time to reach 5.08 m/s=4.16 s
 * t2: substitute values:
 * 0 m/s=5.08 m/s +(-1.22 m/s^2)*t.
 * subtract (-1.22 m/s^2)*t from both sides: (1.22 m/s^2)*t=5.08 m/s
 * divide both sides by 1.22 m/s^2, so t2=time to return to rest=4.16 s
 * t=__time starting and ending at rest=8.32 s__
 * 02. A stone is dropped into a river from a bridge 50 m above the water. Another stone is thrown vertically down 1.00 s after the first stone is dropped. The two stones st rike the water surface at the same time. **
 * i). How long does it take for the first stone to reach the water surface? **
 * use equation x-x0=v0*t+1/2*a*t^2
 * substitute values:
 * 50 m= (0 m/s)*t+1/2(9.8 m/s^2)*t^2
 * divide both sides by 4.9 m/s^2 and get 10.2 s^2=t^2
 * take square root of both sides
 * since time can't be negative, __t=3.19 s.__
 * ii). What is the initial speed of the second stone? **
 * use equation x-x0=v0*t+1/2*a*t^2
 * substitute values:
 * 50 m=v0*(3.19 s-1.00 s)+1/2(9.8 m/s^2)*(3.19 s-1.00 s)^2
 * evaluate to get 50 m=(2.19 s)*v0+(4.9 m/s^2)*(2.19 s)^2 or 50 m=(2.19 s)*v0+23.50 m
 * subtract 23.50 m from both sides to get 26.50 m=(2.19 s)*v0
 * divide both sides by 2.19 s
 * v0=__initial speed of second stone=12.10 m/s__
 * v0=__initial speed of second stone=12.10 m/s__

Professor's Note : Very Good effort (liked the explanations as well !). Do not include numerical values in the final answer -- check my page

__ **Week three (06th of February, 2012)** __

1. If ** B ** is added to ** A ** the result is 6.0 iˆ + 1.0 jˆ. If ** B ** is subtracted from ** A **, the result is - 4.0 iˆ + 7.0 jˆ.

i. Find ** A ** and ** B **
 * let **A**+**B**=**R** and **A**-**B**=**V**
 * **R**x=6.0*i^=**A**x+**B**x
 * **R**y=1.0*j^=**A**y+**B**y
 * **V**x=-4.0*i^=**A**x-**B**x
 * **V**y=7.0*j^=**A**y-**B**y
 * add **R**x+**V**x so that **B**x cancels out and you are left with 2**A**x=2.0i^ and divide both sides by 2 to get **A**x=1.0i^
 * add **R**y+**V**y so that **B**y cancels out and you are left with 2**A**y=8.0j^ and divide both sides by 2 to get **A**y=4.0j^
 * now plug values of **A**x and **A**y back into either **R**x and **R**y or **V**x and **V**y equations to find **B**x and **B**y

ii. Find the magnitudes of ** A ** and ** B **
 * |**A**|=sqrt((**A**x)^2+(**A**y)^2)
 * |**B**|=sqrt((**B**x)^2+(**B**y)^2)

iii. What is the angle between ** A ** and ** B ** Professor's Note : Very well done ! Keep up the good work !!
 * use **A•B**=|**A**|*|**B**|*cos(theta)
 * **A**x***B**x+**A**y***B**y=|**A**|*|**B**|*cos(theta)
 * plug in values of **A**x, **A**y, **B**x, **B**y, |**A**|, and |**B**|
 * theta=arccos((**A**x***B**x+**A**y***B**y)/(|**A**|*|**B**|)

__ **Week four (13 th February, 2012)** __

1. A 0.5 kg rock is projected from the edge of the top of a building with an initial velocity of 10.5 m/s at an angle 54º above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 15.3 m from the base of the building. Assume: The ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s². What is the height of the building?


 * first I have to solve for time. I know x-x0=15.3 m, v0x= 10.5*cos54 m/s, and ax=0, so I plug these values in to x-x0=v0x*t+1/2(t)^2 to get t.
 * then I plug in v0y=10.5*sin54 m/s, and ay=-9.8 m/s^2, and the value I found for t into equation y-y0=v0y*t+1/2(t^2) to find y-y0.

--- No new work was found -- Professor