(II) An 85-m-long train begins uniform acceleration from rest....

Question

(II) An 85-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/s when it passes a railway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Fig. 2–45.)

Accepted Answer

Hey everyone in this problem, a vessel measuring 120 m in length starts its uniform acceleration from a complete stop when the front of the vessel reaches a buoy, located 300 m away from its initial position is traveling at a speed of 12 m per second. And we're asked, what will the speed be of the rear end of the vessel as it passes the buoy? We're given four answer choices all in meters per second. Option A 24 option B 14, option C 20 option D 12. Ok. So let's think about this. We wanna know the speed when the rear end of the vessel passes the buoy and we're told some information about when the front end of the vessel reaches that buoy. Ok. So we have our buoy and then we have our vessel and we're just gonna draw it as kind of a cylinder or sorry, an oval. Hey, this is our vessel and we know that the vessel is 120 m long and that vessel has some speed right now. What we want to find out is the speed when the rear end reaches the buoy. All right. Well, what do we know about this situation that we've drawn here? We know that the speed we not when this vessel is at this position is 12 m per second, we wanna find the final speed. VF ok. And that final speed is going to correspond to when the end reaches that buoy. We know that in order for this to happen, Delta X or the distance traveled must be 120 m and that ship or vessel is 100 and 20 m long. So to go from the front of it being at the buoy, to the back of it being at the buoy, it needs to travel 120 m. We don't know the acceleration of this ship and we don't know how long this takes. Hm So we only have two known values which is not enough to calculate our final velocity that we're looking for. Ok. So we need to use some other information that was given in this problem in order to solve for via. So let's think about the other information we're given, ok. And the other information we're given is that when it's three, it's 300 m away from the initial position, ok? That buoy is 300 m away from the initial position and we know that it starts from rest. So let's draw this up. We have our Bowie again and let's do this in b so we have our Buie and we have our vessel and now the vessel is 300 m. Oh why from that buoy? And this is the initial starting position and we're told that this starts from a complete stop. And so in this case, we know that the initial speed is gonna be 0 m per second. We know the final speed is 12 m per second. Ok. The speed that this vessel is traveling when the front reaches the buoy. And we know that the distance traveled in this case delta X is 300 m. So that's three known values. This is gonna allow us to calculate one of the others. And the one we're gonna calculate is the acceleration. Ok? We're told that this ship has a uniform acceleration. If it has a uniform acceleration, that means that the acceleration from this initial point up until the front of the vessel reaching that buoy is gonna be the exact same as the acceleration from that point until the end of the vessel is reaching the buoy. Ok. So we can find the acceleration using this information we've written down in blue. We are then gonna be able to use that same acceleration for that first part that we wrote down to find that final velocity. All right. So we wanna choose a kinematic equation with VNVF delta X are three known values and the acceleration a the one we're looking for and that is gonna be VF squared is equal to V, not squared plus two multiplied by a multiplied by delta X substituting in our values 12 m per second, all squared is equal to, well, V not is zero. So that first term on the right hand side goes to zero, we have two multiplied by a multiplied by 300 m. Simplifying on both sides 144 m squared per second squared is equal to 600 m multiplied by A and if we divide both sides by 600 m, we get that the acceleration A is 0.24 m per second squared. Hm All right. So now we know the acceleration of this boat. And again, we're told that's a uniform acceleration. So we can go ahead and add that to our information at the top. We now have this acceleration of 0.24 m per second squared. And if we take a look at the information we have there, now we have three known values V not delta X and A, we wanna calculate the final velocity V or the speed that that vessel is traveling when the end of it reaches the boo. All right. So we're gonna choose the exact same equation the one without the time T because we don't have information about that. And that's not what we're looking for. So we have, again, VF squared is equal to V not squared plus two, multiplied by a multiplied by delta X substituting in our values. We get that VF squared is equal to 12 m per second, all squared plus two, multiplied by 0.24 m per second squared, multiplied by 120 m. Simplifying the right hand side, we get that VF squared is equal to 201.6 m squared per second squared. And finally taking the square root and we're gonna take the positive square root because we were asked for the speeds we're looking for just the magnitude. Um So we take the positive route and we get that, the speed is about 14.1986 m per second. All right. And that's what we were looking for. That is the speed of this vessel. When the rear end passes the buoy. If we compare this to our answer choices, we're going round to the nearest meter per second and we can see that the correct answer is going to be b 14 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.

(II) An 85-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/s when it passes a railway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Fig. 2–45.) <IMAGE>

Key Concepts

Uniform Acceleration

Relative Motion

Kinematic Equations

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