alimourad


 * 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? **

v^2=(v0)^2+2a(x-x0)

v=305 m/min*min/60 s= 5.08 m/s

(5.08 m/s)^2=(0 m/s)^2+2(1.22 m/s^2)(x-x0)
 * divide both sides of the equation by (2.44 m/s^2).
 * x-x0= __10.58m__
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? **
 * v=v0+at.
 * 5.08 m/s=(1.22 m/s^2)*t
 * divide both sides by 1.22 m/s^2, 5.08 m/s=4.16 s
 * 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, 4.16s than multiply it by two to get-
 * t= __8.32 seconds__
 * 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? **
 * x-x0=v0*t+1/2*a*t^2
 * 50 m= (0 m/s)*t+1/2(9.8 m/s^2)*t^2
 * square root both sides of the equation
 * t= 3.19 seconds
 * ii). What is the initial speed of the second stone? **
 * x-x0=v0*t+1/2*a*t^2
 * 50 m=v0*(3.19 s-1.00 s)+1/2(9.8 m/s^2)*(3.19 s-1.00 s)^2
 * 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
 * divide both sides by 2.19 s
 * v0= __12.39m/s__
 * divide both sides by 2.19 s
 * v0= __12.39m/s__

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


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