Matthew+Baucum


 * 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? **

Vo = 0m/s, V= 305m/min, A=1.22m/s^2

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

V^2=Vo^2+2a(X-Xo) 5.08^2=0+2(1.22)(X-Xo) 25.81=2.44(X-Xo) 25.81/2.44=(X-Xo) 10.6=(X-Xo) The car reached maximum speed at **10.6m**.


 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? **

Part 1/2 Vo=0m/s, V=5.08m/s A=1.22m/s^2 X-Xo=10.6m 10.6=1/2(0+5.08)t 10.6=2.54*t 10.6/2.54=t t=4.17 4.17 Seconds for part one.

Part 2/2 Vo= 5.08m/s V=0m/s A=1.22m/s^2 X-Xo= 179.4 179.4=1/2(5.08)t 179.4/2.54=t 70.6=t 70.6 seconds for part two 4.17 + 70.6 = **74.77s**

V=0 A= -9.8m/s^2 V-Vo= 50 >> professor's note == given is x-x0 = 50 m, mistake! 50=9.8(t) t=**5.1s**
 * 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? **
 * i). How long does it take for the first stone to reach the water surface? **

V=0 A=-9.8m/s^2 V-Vo= 50 T=4.1s 0=Vo-9.8*4.1 0=Vo-40 40=Vo Vo= **40m/s**
 * ii). What is the initial speed of the second stone? **


 * Professor's Note : Very Good effort - Like to have more explanations and try not to include numerical values in the final answer -- check my page **

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