A 1000 kg weather rocket is launched straight up. The rocket moto...

Question

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. What was the rocket's acceleration during the first 16 s?

Accepted Answer

Hey, everyone. Welcome back in this problem. During a voluntary vertical takeoff, A drone fly of mass 20g produces a constant acceleration by flapping its wings for 1.2 seconds At T equals 1.2 seconds. The drone fly stops flipping its wings And .2 seconds later, it is 3.2 m above the takeoff level and were asked to calculate the acceleration produced during the 1st 1.2 seconds. Alright. So we're told in this problem that we have constant acceleration, we have constant acceleration. We know we can use our you am equations or cinematic equations. Now, in this case, we have to consider two different stages. Okay. We have when the drone fly is flipping its wings And we're going to consider this stage one. What information do we have here? Well, we know that the initial speed, 0m/s. Okay. Starting from rest and then it's taking off. We don't know what the final speed is. The acceleration. A one is actually what we're trying to find the acceleration during this 1st 1.2 seconds, which is when it's flipping its wings. We know that the time it spends here is 1.2 seconds. Okay. It starts after it flips its wings for 1.2 seconds and then stops. The amount of time here is t one equals 1.2 seconds. And delta Y one, we also don't know anything about. So in just this first stage, we have two pieces of information that we know one that we want to find out. Now, two things, two knowns is not enough to solve this problem. Remember with our kid a Matic or um equations, we have to know three things. So we actually don't know How to get a one yet. Let's look at stage two. Stage two is gonna be when it's not flipping its wings. Okay. After it stopped Stage two, we're gonna write the same information. So v not to, We don't know what speed the drone fly is going once it stops flapping its wings, but we do know that this is going to be equal to VF one. Okay. The speed that it's going when it's immediately stops flapping its wings is gonna be the exact same thing, exact same speed as the final speed when it's flipping its wings. Alright. So we don't know the value but we know that there's a relationship there, The final speed, sorry, the F two, we don't have information about the acceleration. Okay. Here it's not flapping its wings. So the only acceleration is going to be due to gravity and that's gonna be negative 9.8 m per second squared. Okay. We're assuming that the upward direction is going to be positive the time it spends In this stage where it's not flipping its wings, we were told .2 seconds later there's some height. Okay. So we're going to say that it spends .2 seconds in this stage And the height. Delta Y two. Well, we're told it's 3.2 m above the takeoff, okay. But above the takeoff level considers delta Y one plus delta Y two. Okay. So we know that the, some of those is going to be 3.2 that tells us that the change in height of this second phase is going to be that 3.2 m minus whatever the change in height was of the first phase. So we have 3.2 m minus delta Y one. So now we have this relationship. V not two is equal to V up one and we have a relationship between delta Y two and delta Y one. What this allows us to do is write an equation for the first stage, okay with VF one and delta Y one is unknown and then write an equation for the second stage where we can also have V F one in delta Y one as unknowns. That's two equations, two unknowns. And we'll be able to solve for those unknown values. Once we have one of those unknown values, then we'll have three known quantities for stage one, which will allow us to calculate a one. All right. So let's start with the first stage. We want the equation without acceleration here, which seems a little bit backwards because we're looking for acceleration. But remember we need to use those quantities that are related between our two stages first. So we're gonna use the equation without acceleration. We have the delta Y one is equal to one half times V, not one plus V F one times T we can substitute in the information we know at this point. All we know is the initial speed V not is zero. So this is going to go to zero and we know the time T, so let's switch over to the other equation for the other equation or for the other stage first. Now, in this stage, we know the acceleration. So we want to use the equation without VF two. If we do that, we have the phone. Yeah, the Delta Y two is equal to V not to times T two plus one half A two T two squirt substituting what we know. We know that delta Y two is equal to 3.2 m minus delta Y one. You can write that here. 3.2 m minus delta Y one is equal to the initial speed V. Not too well, that's equal to V F one. The time is 0.2 seconds plus one half the acceleration negative 9.8 m per second squared times the time, 0.2, We're running out of space right on an angle here, 0.2 seconds squared. Now we're gonna simplify what we can here. On the left hand side, we get 3.2 m minus delta Y one and that's equal to, on the right hand side, we have V F one times 0.2 seconds. We can't do much with that yet. Now, one half times negative 9.8 m per second, squared, times 0.2 seconds squared gives us minus 1.96 m. We have 3.2 m on the left hand side minus 1. m or sorry, point 0.196 m. On the right hand side, we can add that .196 m to the left hand side. We're gonna have 3.3, meters minus delta Y one is equal to VF one Times 0.2 seconds. Now, if we want to isolate for delta Y Y, we can do that and we can write that delta Y one is equal to 3.396 m. We move the delta Y one to the right. So we have to move this VF one times 10.2 seconds to the left, which is gonna make it negative 0.2 seconds times V F one. So we have this equation we filled in the information we know. And now we have delta Y one is equal to some function related to V F one. Now, we can substitute this, let's call this star, Okay. And we can substitute that into our equation for stage one. And if we do that, we have delta Y one on the left hand side, which is now going to be 3.396 m -0.2 2nd Times VF one is equal to one half. V not one was zero. So we just have times VF one times a time, 1.2 seconds. All right, let's move all of the V F one terms to the right hand side together. We have 3.396 m is equal to one half times. 1.2 is gonna give us 0.6 seconds times V F one and then we're gonna add 10.2 seconds times VF one when we move this to the right hand side. And so we end up with 0.8 seconds times V F one, we can divide by 0.8 seconds to get VF one. We get that V F one is equal to 4.245 m per second. Okay. So we have this value 4.245 m per second. But remember that's not what we're trying to find. We're trying to find the acceleration in this first stage, the acceleration in the 1st 1.2 seconds, we found VF one, which means that we now have three known values. We know the initial speed, the final speed in the time with three known values in our kingdom attic equations we can solve for the fourth. So now we can choose the equation with acceleration. We're going to have the equation without delta Y one Because we don't know that quantity. We could find it by substituting the F1 back in, but we don't need to. And so the equation for acceleration, we're gonna have VF- one Is equal to V, not one plus 80. We know that V F one is 4.245 m per second. V not one is zero m per second. The acceleration is what we're trying to find and our time T is 1.2 seconds. Okay. So we take our speed 4.245 m per second. We divide it by 1.2 seconds and we get an acceleration of 3.5375 m per second squared. And that is the acceleration of the drone fly during the 1st 1. seconds when it's flapping its wings. We go back up to our answer choices. We see that this corresponds with answer choice. C sorry answer. Choice. B the acceleration is 3.5 m per second squared. Thanks everyone for watching. I hope this video helped you in the next one.

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. What was the rocket's acceleration during the first 16 s?

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