(III) A 360-kg four-wheel flat cart 15 m long is moving horizo...

Question

(III) A 360-kg four-wheel flat cart 15 m long is moving horizontally with a speed of 6.0 m/s. A 110-kg worker starts walking from one end of the cart to the other in the direction of motion, with speed 2.0 m/s with respect to the cart. In the time it takes for him to reach the other end, how far has the cart moved?

Accepted Answer

Hey everyone in this problem, a 115 kg person is stationary at the back of a 9.1 multiplied by 10 to the exponent 3 kg bus moving forward horizontally at a speed of 5 m per second. Suddenly he starts moving toward the front of the bus with a speed of 1 m per second relative to the bus. Given that the bus is 1.4 multiplied by 10 to the exponent 1 m long. We're asked to determine how much the bus moves and the time it takes for the person to reach the front of the bus. We are given four answer choices all in meters. Option A 56 option B 69.8 option C 70 option D 84. The first thing we're gonna do here is figure out how long it takes this person to reach the front of the bus. And we wanna know how far the bus travels in that amount of time. Oh, what's that amount of time? Ok. So we are going to find time it takes to reach the front of the bus overall, but the speed V is equal to the distance divided by the time t we know that this person is moving with a speed of 1 m per second relative to the bus. Ok. So the speed is gonna be 1 m per second. The bus is 1.4 multiplied by 10 to the exponent 1 m long. So to start at the back of the bus and end up at the front of the bus, they're gonna have to travel 1.4 multiplied by 10 to the exponent 1 m. And that's divided by that time, we're looking for tea. And so our time T is gonna end up being equal to 14 m divided by 1 m per second, which gives us 14 seconds. All right. So we're gonna put a red box around that. We're gonna need that value in just a minute. So we know that it's gonna take 14 seconds for him to reach the front of the bus. So we wanna know how much the bus is gonna move in 14 seconds, right? So now we wanna find, let's do this in a different color. We wanna find the distance the bus travels in that amount of time. So in that 14 seconds, now we're gonna use the same equation. The speed V is equal to the distance divided by the time T. So our distance is gonna be V multiplied by T but we have to be a little bit careful here. OK? The speed that we're looking for is the speed of the bus relative to the ground? Ok. And that's not the same as the speed we were given in the problem. Hm So initially the person is stationary and the bus has a speed of 5 m per second. Once the person starts moving, that's gonna affect the speed of the bus. They recall that that combined momentum of the person in the bus is conserved. So if the momentum of the person increases because their speed increases from zero to now 1 m per second, that means that the momentum of the bus must decrease in order for momentum to be conserved, the mass of the bus is not changing. And so the speed of the bus must decrease slightly in order to compensate, right? So the speed that the bus is gonna be traveling once this person starts walking is not exactly 5 m per second. So we actually have to calculate that speed. VBG first before we can find this distance. Now I mentioned we have this conservation of momentum. We can use that to find this speed or we can think about the speed of the center of mass. And we know that the velocity, the center of mass of this bus is going to remain constant. So what we can do is think about that velocity, right. So if we think about the velocity, the center of mass of this bus person system, we know that this is gonna be equal to the mass of the person multiplied by their velocity relative to the ground plus the mass of the bus multiplied by its velocity relative to the ground all divided by the sum of the masses. So we have M PV PG plus MBVBG divided by MP plus MB. Now, the value we're looking for right now is this VBG value that's that velocity of the bus relative to the ground that we need, hey, we know the masses of the boss and the person. So we need to think about this velocity V of the person with respect to the ground. OK. So we need in our calculation VPG but we were given the velocity of the person with respect to the bus. And so recall that the velocity of the person with respect to the ground is equal to the velocity of the person with respect to the bus plus the velocity of the bus with respect to the ground. OK? So this middle terms match up and the outside terms give us that final result. Now the velocity of the person with respect to the bus we're told is 1 m per second. And then the velocity of the bus with respect to the ground is what we are looking for. OK. So we can substitute this VPG value into our equation for this velocity of the center of math. Now the velocity of the center of mass is that 5 m per second. That initial velocity that we're told. And so we know that value as well. So let's try to simplify and rearrange what we have here first before we substitute in values. So we have that VCM is going to be equal to MP multiplied by VPG, which we now know we can ride as 1 m per second, less that speed of the bus with respect to the ground. What MB VBG all divided by the sum of the masses MP plus MP, we can multiply both sides by the sum of those masses and then expand and rearrange in that numerator. So we get VCM multiplied by MP plus MB. That's gonna be equal to MP multiplied by 1 m per second. The other two terms we have are both gonna have this B uh VBG term. So the velocity of the bus with respect to the ground. So we can factor that out. So we get plus VBG multiplied by that sum of masses MP plus MB. Now we want to solve for VBG. So we're gonna subtract MP times 1 m per second from both sides. We're gonna be left with VBG. OK. We're rearranging multiplied by the sum of masses MP plus MB is equal to VCM multiplied by MP plus MB minus MP multiplied by 1 m per second. Final thing we need to do is divide by that sum of masses. So we get VVG, the velocity of the bus with respect to the ground is gonna be VCM. OK. The sum of masses will divide out in that term minus MP, multiplied by 1 m per second, divided by the sum of masses MP plus MP. OK. So now we have this expression for the speed of the bus with respect to the ground that we want, we can go ahead and substitute in our values. And so we get that VBG is going to be equal to 5 m per second minus the mass of the person 115 kg multiplied by 1 m per second, divided by the sum of the masses. So the mass of the person 115 kg plus the mass of the bus 9.1 multiplied by 10 to the exponent 3 kg. And when we work all of this out, we get that the velocity of the bus relative to the ground is going to be approximately equal to 4.9875 m per second. So it has not changed very much, but it has decreased a little bit. And the decrease makes sense. Remember we said the momentum of the person is increasing. So the momentum of the bo must decrease a little bit in order to keep that momentum concerned. All right. So we have the speed of the bus relative to the ground. We can get back to this calculation in green that we were trying to do. We want to find how far this bus traveled in that 14 seconds, it took for the person to walk from the back of the bus to the front of the bus, we now have the correct speed to use. So the distance traveled D is gonna be equal to 4.9875 m per second, multiplied by 14 seconds, which gives us a value of approximately equal to 69.825 m. So that is the distance the bus will travel as this person walks to the front. If we take a look at our answer choices, we round to one decimal place, we can see that the correct answer is going to be option B 69.8 m. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Relative Motion

Conservation of Momentum

Kinematics

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