Geoff+Maynard

First write down the given data: a=1.22m/s v=305m/min Δ X=190m v0=0m/min Convert needed to SI units: 305m/min->5.083m/s
 * Week 7 : visited no new work -- Professor**
 * 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. **
 * 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. **

Find the corresponding equation needed to solve for the distance: v^2=v0^2+2a(Δ X) Plug in the data and solve for Δ X Δ X=(v^2-v0^2)/2a =(5.083^2-0)/(2(1.22)) =10.59m You need to break the problem up into 3parts: (1)acceleration, (2)constant, and (3) deceleration We know the distance for the car to accelerate and decelerate, so we need to find the constant speed distance: 190-10.59(2)=168.82m Solve each part for time: (1) Δ X=1/2(v0+v)t (2) Δ X=vt-1/2at^2 (3) Δ X=1/2(v0-v)t 10.59=1/2(5.083)t 168.82=5.083t 10.59=1/2(5.083)t t1=4.16s t2=33.21s t3=4.16s
 * 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? **

t1+t2+t3=41.53s


 * professor's note : it only has an acceleration & a deceleration **

First write down the given data: a=9.8m/s Δy =50 v0(first stone)=0 (2nd stone) dropped 1sec later
 * 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. **

Find the equation that will solve for time:
 * i). How long does it take for the first stone to reach the water surface? **

Δy =v0t+1/2(a)t^2 Plug in the data and solve for t: 50=1/2(9.8)t^2 t=(100/9.8)^(1/2) t=3.19s

Find the equation to solve for initial speed: Δy =v0t+1/2(a)t^2 Plug in the data: since the first stone took 3.19s to fall, the second will be 2.19s 50=v0(2.19)+1/2(9.8)(2.19)^2 26.4991=v0(2.19) v0=12.1s
 * ii). What is the initial speed of the second stone? **


 * Professor's Note : Good effort - Like the explanations !! Do not include numerical values in the final answer -- check my page **


 * VISITED THE PAGE, NO NEW WORK WAS FOUND --- PROF.**

Week 4: State the given data, which includes; Vo, theta, (X-Xi), gravity(everyone should know this already) Now solve for t using the equation: (X-Xi)=(Voxcos(theta))t Plug in the given data to solve for t then find (Y-Yi) using the equation: (Y-Yi)=Voycos(theta)-1/2gt^2 Plug in the data and you will find (Y-Yi), it will be a negative number, just switch it to positive for building height- this is due to the way the problem is stated.
 * 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? **

**Professor's Note: good job! , also visit my page for more hints for tomorrow Quiz**