Quacy-c

Professor's Note : Good effort - but incomplete!-- check my page
 * 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? **
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? **
 * 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? **
 * ii). What is the initial speed of the second stone? **
 * Answer key **
 * For part 1 use **
 * -the equation V^2=Vo^2+2a(X-Xo) **
 * - change 305m/min in to SI unit(305/60)=V **
 * - sub the values on **** V^2=Vo^2+2a(X-Xo), then solve for change in X **
 * - sub the values on **** V^2=Vo^2+2a(X-Xo), then solve for change in X **

__ **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 ** ii. Find the magnitudes of ** A ** and ** B ** iii. What is the angle between ** A ** and ** B ** __** Answers key **__ //**First draw the diagram for the information given**// //**(i)Subtract the result from the second from the first result to find B. The A will be obvious.**// //**(ii) To find the magnitude you use the Sin(x) to find the x magnitude ,and then use the Cos(x) to find the Y magnitude.**// //**(iii) Then use Tan(x) to find the angle between A and B.**//
 * For part 2 use **

Professor Note : I can't agree with part ii and iii, please check my page

__ **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?

Answer key - First thing to do is draw the diagram - use the formula detaX= VoX*t + 1/2 to solve for t, since we know deta x = 15.3, V0x = 10.5cos 54 ax=0 - after solving from t ,we know have all the necessary values to solve for deta Y which is the height, so we can use the equation deta Y =Voy*t+1/2(t^2)