Physics gr11 review (acceleration)

AI Thread Summary
The discussion revolves around solving physics problems related to acceleration and motion. A train accelerates uniformly, and the speed of the last car as it passes a worker is calculated to be approximately 41.67 m/s. The second problem regarding a hot air balloon and a dropped can is noted as confusing, with a request for clarification on the data provided. The third problem involves two stones dropped at different times, with calculations showing they are 22.1 meters apart when the second stone reaches a speed of 23 m/s. Finally, the height of a cliff based on the sound delay is incorrectly calculated, indicating a need for further verification of the solution.
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1. The problem statements, all variables and given/known data

1. a 100m long train accelerates uniformly from rest. If the front of the train passes a railway worker 150m down the track at a speed of 25m/s, what will be the speed of the last car as it passes the worker?

2. a hot air baloon is rising up with a constant velocity of 4m/s. As the balloon reaches a height of 4m above the ground the balloonist accidentally drops a can of pop over the edge of the basket. How long does it take the pop can to reach the ground?

3.a stone is dropped from the roof of a high building. a second stone is droped 1s later. How far apart are the stones when the second one reached the speed of 23m/s?

3. a rock is dropped from a seacliff and the sound of it stirking the ocean is heard 3s later. If the speed of sound is 330m/s, how high is the cliff?



Homework Equations


These are the equations we were given to solve the questions:
(it might look messy, f is final and t is time)
Vf^2=V1+2ad
Vf=V1+at
d=V1+1/2at^2
d=Vft-1/2at^2
d=(V1+Vf/2)t

honestly i tried doing these, but i just don't get it!
 
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1. 150/25 = 250/x
x = 41.6666667m/s

2. this one is weird. can't help. are you sure you copied all the data?

3. vf^2=vi+2ad
23^2=0+2(9.8)d
d=27 m (stone 1)

d = (vi+vf/2)t
27 = (0+ 27/2)t
t=2 seconds( the first stone reached 27m in two seconds)

2s -1s = 1s so we want to calculate the position of stone 2 at one seconds so

d = vi + .5at^2
d = 0 + .5*9.8*1^2
d = 4.9m

21 - 4.9 = 22.1m difference

4. 3s * 330m / 1s = 990 m

have no clue if any of these are correct. please check my work guys
 
Hey thanks for the help, the first and third make sense.. yea it know the second is weird but i read the question over and that's what it says. the 4th one is way off lol, it was a pretty long solution that came to a heights of h=4.9(2.9)^2
 
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