A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and land at ground level on the other side. He has done this stunt many times and approaches it with confidence. Unfortunately, the motorcycle engine dies just as he starts up the ramp. He is going 11 m/s at that instant, and the rolling friction of his rubber tires (coefficient 0.02) is not negligible. Does he survive, or does he become crocodile food? Justify your answer by calculating the distance he travels through the air after leaving the end of the ramp.

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Accepted Answer

Hey, everyone in this problem, we have a film shoot and there's a stunt scene where the hero drives his car up a 25 degree incline slope. So let's go ahead and draw that slope out. We have a 20 degree inclined slope leading to a 4 m high cliff or cliff is 4 m high. The car is going to jump off that cliff and cross a fire that is 9 m on the ground. So 9 m of fire and hopefully reach the ground safely, hang on the other side of the. So the hero starts up the slope with a speed of 12 m per second. Ok. So we're just gonna draw this stunt driver as a dot And their initial speed which we're gonna call VNO is gonna be 12 m per second. The engine is going to get locked. We have rolling friction acting because we have a coefficient of friction, 0.025. And we're asked to determine whether he lands safely on the ground falls in the fire. We're also told that their flight time is 1.3 seconds. So we have four answer choices here. Option A is the car does not cross the cliff at all. Ok. So it doesn't even make it up. Option B is that the car lands safely on the ground? Option C is that the car is going to fall into the fire? And option D is that we can't determine it from the information we're given. All right. So we have our car starting at the bottom. It's gonna move up our incline and then it's gonna take some flight path and maybe it'll land on the other side of the fire, maybe it'll land in the fire. That's what we have to figure out. Now, if we think about just the flight part of this. Ok. So if we think about when the car is on the edge of the cliff and then they start to do this flight portion, what do we know? Ok. Well, we don't know the speed. We don't know the distance that they're traveling. That's what we were trying to find out. We don't know a whole lot. Ok. So in order to solve this problem, we need to figure out some more information. And one of the things that we can figure out is the speed at the edge of the cliff, say that initial speed for their flight, which is gonna be the final speed if we look at the motion of the ring. Ok. So we want to find the speed at the end of the ramp. Well, we know the initial speed, we can find how long we travel. OK? Based on knowing the height and the angle, but we don't know anything else. OK. So again, we need more information. Remember that with our kinematic or U AM equations, we need at least three known values. And so we're gonna have to start all the way back at the beginning and find the acceleration of this car along the R. OK. So step one is going to be to find the acceleration. Oh Wow. Right. So let's get to it, recall that we can relate the acceleration to the forces. And so in this case, the some of the forces, we're gonna say that our X direction is parallel to our recline. It's gonna be equal to the mass multiplied by the acceleration. And in the one direction, the sum of the forces is just going to be equal to zero. And this car is not moving up and down perpendicularly to the incline and staying on that incline and moving side to side, let's go ahead and draw a free body diagram. We have the normal force acting upwards perpendicular to our incline. The force of gravity is going to act straight downward. So we can break that up into its components. So we have what we're gonna call the Y component that's going to act perpendicularly to the incline. Then we have the X component acting parallel to the incline and we also have our force of friction and that's kinetic friction. So we're gonna call that FK and that's acting in the same direction of the X component of the force of gravity. OK. Along that incline to the left. Now, these look, let me make that a little bit more parable. There we go. Yeah, that's gonna oppose the motion. The force of friction opposes the motion, we're gonna say up and to the right on our tilted axis is our positive direction. So if we get back to our sum of forces equations, the sum of forces in the X direction when we have this force of kinetic friction and the X component of the force of gravity both acting in the negative direction. And so we have minus FK minus FGX is gonna be equal to M A. Now the force of kinetic friction, her call is going to be mu K multiplied by the force of gravity is given by M multiplied by G. We want the X component. So we're looking at sine here. So we have minus UK N minus mg sinus theta is equal to the mass um multiplied by the acceleration A X. Now we can see these mass terms, you might be thinking we aren't given the mass. So maybe we can't figure this question out. Let's keep going and you'll see why we can keep going without the mask. OK. Now we don't know the normal force yet. So that's one thing we need to figure out before we can solve for this acceleration that we're looking for. So let's go over to our sum of forces in the Y direction by normal force. In the Y direction, we have N acting in a positive direction. We have the Y component of the force of gravity acting in the negative direction. So N minus Fgy is equal to zero. That tells us that the normal force is gonna be equal to M multiplied by G multiplied by cosine of the. All right. Going back to our X direction, we're gonna substitute in that normal horse. We get minus in the UK multiplied by M multiplied by G multiplied by post, the minus M multiplied by G multiplied by sine data is equal to M multiplied by A X. And what you can see is that we have the mass M in every single term. So we can divide our equation by M. So we divide it and we don't have to worry about it. OK? So our acceleration A X is then going to be negative that coefficient of friction, 0.025 multiplied by G 9.8 m per second squared multiplied by cosine 25 degrees minus 9.8 m per second squared multiplied by sign of 25 degrees. And if you work all of that out, you're gonna get an acceleration of about negative 4.3637 m per second squared. All right. So we have this acceleration. Remember, we're trying to figure out if this car is going to land safely or if it's going to fall in the fire, we have our acceleration along the ramp which is going to now allow us to find the speed at the end of the ring. And we need that in order to figure out how far this thing is going to why off the rip off the cliff. Ok. So step two, find speed at the end of the rant for the. Now, we can go ahead and use our kinematic equations now. OK. So the initial speed V not along that ramp, we're told is 12 m per second. We wanna find the final speed via, we now know that the acceleration is negative 4.3637 meters per second squared. And delta X the distance traveled or the displacement, let's go back up to our diagram. We know that when we look at our angle of 20 degrees and sorry, this should be 25 degrees, 25 degrees. When we look at that angle, the opposite side is 4 m and we wanna find the length of the hypo. And so we can use are trigonometric ratios in right delta X. The distance of this car travels as four divided by sine of 25 degrees. Uh Now we have three values that we know one unknown. We're gonna choose a kinematic equation with these four variables. We have VF squared is equal to V not squared plus two A delta X substituting in our values VF squared is equal to 12 m per second. All squared plus two, multiplied by negative 4.3637 m per second squared multiplied by four divided by S 25 degrees. If we work all of this out, we get that the final speed. The app after taking the square root is going to be 7.8356 m per second. And we know that that's a positive velocity. We chose the positive route because it's moving to the right that direction that we've chosen to be positive. All right, so we have this final speed coming off of the ramp. Let me put that in red. Now, we can get to the real nitty gritty of this question answering the question itself. What is the flight? How far does this car go? Is it going to make it safely through the fire? And we are going to for step three, write this in the loop. Ok. So everything in black pertains to while the car is on the hill or on the cliff, everything in blue is going to be during this flight. Ok. So coming off the ramp, the initial velocity in the X direction, it's going to be that final velocity we just found of 7.8356 m per second multiplied by cosine of 25 degrees. OK? Because the ramp is on an incline. So this final velocity is also on the incline. When we switch to looking post clip, we wanted to switch back to our regular axes, X and Y up and down. OK. So we wanna kind of come off of that incline and look at our regular X and Y directions. What about our final X velocity? And well, it's just gonna be the initial X velocity, we gonna assume we have no air resistance. So the acceleration is going to be zero in the X direction. So we really only have three variables to worry about V, not delta X, which we're looking for. And the time T which we're told is 1.3 seconds. So we call that the speed V is equal to the distance divided by the time. In this case, we can write it as V not X is equal to delta X divided by T. So delta X is gonna be equal to V, not X multiplied by T which is equal to 7.8 3 5 6 m per second multiplied by cosine of 25 degrees multiplied by 1.3 seconds. OK? And this gives us a final delta X value of 9.232 m. All right. So we have the distance traveled after the car leaves the cliff. OK. So let's go back up and think about what the question is asking. And remember this horizontal distance of 9.232 m. Going back to our picture, going back to the question, we are told that the fire is 9 m and we're going 9.232 m, which means we are going to land on the other side of the fire safely. Ok. So the correct answer here is going to be option B we land safely on the ground. Thanks everyone for watching. I hope this video helped you in the next one.

Verified Solution

Key Concepts

Kinematics

Projectile Motion

Energy Conservation