kevin+l

__**week one**__ 1) given values total distance=delta x=190m Max V= 305 meters/ minute Acceleration =1.22meters/second^2 i) The distance needed to achieve max velocity can be found using the rate of acceleration and the max speed. The max speed first needs to be converted from meters/minute to meters/second 305m/min*(1min/60seconds)=5.08 meters/second= max velocity. From there I used V^2=Vo^2+2a(delta x) the first v being the max velocity of 5.08m/s and the Vo being the initial velocity in this scenario being zero. (5.08m/s)^2=0^2+2(1.22m/s^2)(delta x). ((5.08m/s)^2)/((2)(1.22m/s^2))=(delta x)= 10.576meters

ii) The time it takes to start and stop the car added to the time traveling full speed will equal the total time spent while traveling the full 190m. To first find the starting time required we have the formula v=vo+at where plugging everything in that was found above looking to find t 5.08m/s=0+1.22m/s^2t so 5.08/1.22=t=4.164s. This is multiplied by two because the car also has to stop leaving the start and stop time in total to equal 8.328 seconds. It took 10.576 meters to start the car and another 10.576 to stop it subtract the two from the full distance leaving 168.848 meters for the car to travel at max velocity of 5.08 until it is required to slow down. So 168.848m/5.08(m/s)=33.23seconds. this is added to the start and stop 8.328+33.23= = time for car to travel the full 190 meters= 41.558 seconds

Professor's Note : VERY Good effort - Like the explanations a lot !! avoid numerical values in the final answer -- check my page

where is the answer to # 2?


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

__**week three**__ 1. A 0.5 kg rock initial velocity of 10.5 m/s angle 54º above the horizontal Due to gravity, the rock strikes the ground at a horizontal distance of 15.3 m The acceleration of gravity is 9.8 m/s². What is the height of the building?

i) if .5 (the mass of the rock)is multiplied to the cosine of 54 (** Very wrong idea -- please check the comment below **) is taken this gives you the velocity in the x direction. if.5 (the mass of the rock)is multiplied to the sine of 54 would result in the y direction. nothing has to be done to the number 54 because it is assumed to be in terms of x in the positive direction.

ii) in order to find the height of the building first the time for the rock to travel in the x direction according to the x velocity and distance in the x direction. with out air resistance the rock travels in the x direction at a constant rate (velocity in the x) throughout the duration of the rocks flight.

iii) then the time taken to travel the distance horizontally is also equal to the time it it takes for the rock to rise up and go back down to hit the ground . using the initial velocity in the y direction and knowing that at max height the velocity is zero then time it take for the rock to rise and fall to the same point can be found. this is subtracted from the time previously found for the rock to travel in the x direction.

iv) the velocity used to rise is also the final velocity when it falls back to the same point it left from. this means the initial velocity is -10.5. this in combination with the already know variables of acceleration and recently found time the final velocity can be found and from their the displacement of the rock leaving the building to when the rock finally hits the ground can be found.

**Professor's Note: good effort,** **pay attention to the comment in red --> mass is irrelevant, take the magnitude of the initial velocity and multiply by sin theta or cosine theta to find V0y or V0x.** **also visit my page for more hints for tomorrow Quiz**

__**Week 05 ( 20th February, 2012)**__

pelicans intial speed = Vo x-x = 8 m a for the y = -9.81 y-y (a) = 5.4 y-y (b) =2.7

a) initial speed for the y= 0 an the initail x has all the speed. inorder to find Vo which is = to Vx the time in the y direction can an has to be found. to find time Vf has to be found. Vf^2 =(inital v )^2 +2(9.81)(5.4) Vf= 10.3 with vf time can be found with vf= vo +at 10.3= 0 +9.81*t T=1.05 seconds with the time Vx can be found through x-x= VoxT+.5(0)T^2 (a = o in the x direction) 8/1.05=V0x =7.63m/s Vx with a Vy of zero V=Vx initial speed of the pelican = 7.63 m/s

b) now that we have Vx all we need to find is the time it spends in the air. 2.7=(0) T+ .5*9.81*T^2 T= .742 seconds with time and Vx the distance can be found keeping in mind that acceleration in the x direction = zero (x-x)= 7.63*.742= 5.66 meters

No new week was found -- professor -- will visit again !


 * Now that all of you have taken the midterm two, here goes the Quiz-help problem **

Week 10 (April 16th, 2012)

01**. A steel ball of mass 0.5 kg is fastened to a cord that is 70 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of it's path the ball strikes, a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic.

Find,

a) the speed of the ball

b) the speed of the block, both just after the collision

Hint: first draw a diagram. find the velocity of the ball, just before it hits the block.

a. 7 cm is first converted to meters v^2=Vo^2+2(9.8).07 giving an final velocity close to 1.17

With the final velocity the force can be found by multiplying the speed by the weight of the moving object. the final speed of the pendulum with a weight in the x direction is equal to the velocity of the object as if it had fallen straight down the energy in the y direction being completely transferred into the x direction due to centripetal motion and conservation of energy, This is all to find the speed of the ball in the x direction just before the contact with the block.

b. the ball weighing .5 and moving 1.17m/s colliding with a stationary block weighing 2.5. The ball transfers energy in the form of movement / velocity but due to the fact that it is elastic means the two objects colloid and continue to move in the different direction at different velocity (the ball bouncing of the block and returning in the direction it came from. this gives the energy of the system which remains constant but the individual variables change(1/2(m1i)(v1i)^2)+(1/2(m2i)(v2i)^2)=(1/2(m1f)(v1f)^2)+(1/2(m2f)(v2f)^2)

The ball with a weight of .5ends with a velocity of .0975m/s in the opposite direction it original was traveling. while the block at 2.5 kg is now moving .4875 m/s after the collision.