A cannon on a flat railroad car travels to the east with its barrel tilted 30° above horizontal. It fires a cannonball at 50 m/s. At t = 0 s , the car, starting from rest, begins to accelerate to the east at 2.0 m/s². At what time should the cannon be fired to hit a target on the tracks that is 400 m to the east of the car's initial position? Assume that the cannonball is fired from ground level.

Video transcript

Hey, everyone in this problem, we have a daredevil event where an Archer is shooting an arrow while riding a bike. The arrow shot with a speed of 40 m per second and a horizontal acceleration of 0.96 m per second squared at an angle of 25 degrees with respect to the X axis. The bike is initially at rest and then moves along the positive X axis at an acceleration of 3 m per second squared. And we're asked to determine the exact instant at which the arrow must be shot in order to reach a target 250 m east of the Archer's starting location. We're given four answer choices all in seconds. Option A 8.9 option B 10.0 option C 3.5 and option D 12.0. So let's go ahead and draw what we have. So we have the bike and I'm just gonna draw that as a circle and the person on the bike, the daredevil is going to shoot an arrow with the speed of 40 meters per second at an angle of 25 degrees with the horizontal. And then we have our biker who is traveling with an acceleration of 3 m per second squared in that same direction and we want the arrow to reach a 0.200 and 50 m away. Now we're gonna take up into the right to be our positive directions. Now, we have two kind of sets of motion to worry about. We have the bike or the archer itself and then we have the air, ok. So let's start with the bike or the Archer, maybe they're kind of a package. Now, we're told that they start from rest. Bike is initially at rest. So the initial speed V not be of the bike is going to be 0 m per second. We don't know the final speed. We know that they are accelerating at 3 m per second squared. We don't know how far they travel while the arrow is in the, flying through the air and we don't know the time that this takes. And so we have two known values there not enough to really solve for anything. And now we're gonna switch over to the air and look at what's happening with the air. Now, for the arrow, we have two directions of motion to consider we have the X direction and we have the wide direction. Let's just put these cool also together. So we have some more space to work. All right, we have the X direction and the Y direction of the arrow. Now in the X direction, we have some initial velocity. And we can find that by breaking up our velocity of 40 m per second into components using the angle of 25 degrees, the horizontal direction is gonna be the adjacent. So we're gonna relate that through cosine, we have 40 meters per second multiplied by cosine of 25 degrees. And similarly, the not Y A. So the initial velocity in the Y direction is going to be 40 m per second multiplied by s of 25 degrees. All right. Now, the final velocity we don't know. So V fa and VFY A, those are unknown to us, we know that we have a horizontal acceleration a of the arrow of 0.96 meters per second squared. And the acceleration in the vertical direction is gonna be the acceleration due to gravity that acts downward. So it's gonna be negative 9.8 m per second squared K are acting in the opposite direction as what we've chosen to be positive. And now we have our horizontal distances. So we have delta X A and we have delta Y A. Now delta Y A is going to be 0 m. OK. That's our vertical displacement. The arrow once it's shot is going to land at the same height. OK. That's an assumption that we're gonna make. So it's gonna land at the same height as it's shot from. And so that vertical displacement is going to be 0 m. Now for this horizontal value delta X A and let's think about this, we're gonna start, the biker is going to bike some distance and then shoot the air. Yes. So the distance that the biker travels plus the distance that the arrow travels is gonna be this entire distance. Ok. So the biker is gonna travel some distance. We're gonna call that XB and then shoot the arrow and the remaining distance covered is going to be the distance, the horizontal displacement or distance covered by the air. OK. So what that means is we can write delta X A as 250 m minus delta XB. OK. Because the two sum to 250. So now we have a relationship between the variables we're using for the arrow and the variables we're using for the bike and the time for the arrow. We don't know how long this takes to fly. All right. So what we're really looking for is TB and that's the amount of time the biker spends biking before they shoot the arrow. OK. So that's the instant the arrow must be shot. So that's the value we're looking for. Now, in order to solve for that, we need another variable. No. OK. We need three known values. Now, we have just made a relationship between delta XB and our air. So if we can find delta XB, we will be able to calculate the time TB. We need So looking now at our arrow in order to calculate D delta XB, again, we have two known values, we need an additional known value to calculate that. So we skip over to our Y direction, we know three values, we can calculate the time that the arrows flying T A. Then we can use that in the X direction to calculate delta XB and take that back to our biker or archer to calculate TB the value we actually want. Right? So this is why it's really important to write down all of the information you have at the beginning so that you can figure out what you need and the steps you need to take to get there. OK? So we're gonna start in the Y direction finding that time T A. OK? So let's start there. We're gonna choose an equation that has V not A delta Y and T and that equation is gonna be delta Y and in this case, delta Y A is equal to V, not Y A multiplied by T A plus one half A A T A squared. And that should be a why? All right, substituting in our values zero is equal to 40 m per second, multiplied by sign of 25 degrees multiplied by T A plus one half multiplied by negative 9.8 meters per second squared, multiplied by T A squared. We can factor out one of these T A values. OK. So we factor out T A to simplify and we're left with 40 m per second, multiplied by sign of 25 degrees minus 4.9 m per second squared, multiplied by T A. We have two things multiplied together, equaling zero. OK. So if either of these factors are zero, the entire equation will be zero as we want. So we either get T A is equal to zero seconds. Well, that's just the initial time point. OK, when this arrow is being shot. So that's not the time we want, but we can also have the second factor equal to zero. OK. So 40 m per second multiplied by sine of 25 degrees minus 4.9 m per second squared, multiplied by T A equal to zero. Now, if we move this 4.9 m per second squared, multiplied by T A to the right hand side, divide everything by 4.9 m per second squared. We are gonna get that the time that the arrow spends in the air is going to be 3.45 seconds, right? This is a value that we've calculated that we are going to need. We're gonna put a red box around that. So we don't mix it up and then we are going to add it into our information. I'm gonna add it in red just so it's clear that this is a value that we calculated. Not a value that we were given. All right. So now, we can look at our X direction, we have three known values V, not A and T, we can use those to calculate delta XB, which is inside of this delta X A. OK. So we're gonna choose again the same equation because we're working with the same four variables. We're doing this in green once again because we are still dealing with the arrow itself and not the archer. And we are now looking for delta X A and delta XB. OK. So the equation is gonna have delta X A initially, but we've related that to delta XB. So we're really gonna calculate delta XB. But initially our equation will be for delta X A. OK. So we have delta X A is equal to V not a multiplied by T A plus one half A A C A squared substituting in our values. Delta X A. Well, that's just 250 m minus delta XB. This is gonna be equal to 40 m per second, multiplied by cosine of 25 degrees multiplied by 3.45 seconds plus one half, multiplied by 0.96 m per second squared, multiplied by 3.45 seconds, all squared. Yeah, we're gonna move our delta XB to the right hand side, subtract everything else over to the left hand side. And what we get is that delta XB is going to be equal to 250 m minus all of this. We're gonna use a big square bracket. We do not wanna mix anything up. We wanna make sure that negative is applied to the entire turn. So 40 m per second multiplied by cosine of 25 degrees, multiplied by 3.45 seconds plus one half, multiplied by 0.96 m per second. Squared, multiplied by 3.45 seconds all squared. And that is all in that big square bracket right. Now, if we work all of this out on our calculator, be very careful with those brackets. We are gonna get the Delta XB is equal to about 119.216 m. Ok. So that is the distance the bike, the archer travel before shooting the arrow. Ok. Let's go back to our diagram just to make that very, very clear. The bike is going to travel this distance that we just calculated up a 119 point something meters, then it's going to shoot the arrow and the remaining distance traveled is going to be by that arrow. Ok. So delta XB 119 0.216 m. We're writing that in red. So it's clear that we calculated it. Now, when we're looking at our bike or archer information, we have three known values. We wanna find the time TB, we can do that again. We're using that same equation because we're working with the same four variables. So let's go ahead and do that. Let's just find a good place to do that. We're gonna do that on the right hand side here. And so now we're left with finding PB again, the same equation delta XB is equal to V not B PB. First one, A B TV score. Now Delta XB, we just calculated 119.216 m. The initial velocity is zero. So the first term on the right hand side is just zero and then we get one half multiplied by the acceleration of the bike 3 m per second squared, multiplied by the time TB squared, dividing both sides by one half multiplied by three. We get TV squared is equal to about 79.47733 seconds squared. OK? The unit of meters divides up and when we take the square root, we get that the time TB is going to be either plus or minus 8.915 seconds. Hey, we're talking about time. So we want the positive value. So we're gonna go ahead and take the positive route of 8.915 seconds. And that is going to be our answer, ok? So that is the amount of time the biker bikes before shooting the arrow in order for it to reach that 250 m mark. If we compare this to our answer choices, we can see that this corresponds with answer choice a 8.9 seconds. Thanks everyone for watching. I hope this video helped see you in the next one.