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Problem 113 Hard Difficulty

A duck has a mass of 2.5 $\mathrm{kg}$ . As the duck paddles, a force of 0.10 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.20 $\mathrm{N}$ in a direction of $52^{\circ}$ south of east. When these forces begin to act, the velocity of the duck is 0.11 $\mathrm{m} / \mathrm{s}$ in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 3.0 s while the forces are acting.


$0.78 \mathrm{m}$ at $21^{\circ}$ south of east


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Video Transcript

we begin this question that calculating the acceleration off the duck. For that we have to use Newton's second law on I'm using the following reference frame. These is my Y axis and disease, my ex access Then Newton's second law tells us that the X direction the net force is equals to the mass off the look times its acceleration, the X direction. The Net force that acts on the duck index direction is composed by true forces half one, which points to the positive direction and the X component off after two, which also going to the positive X direction and these East Coast with a mass off the dark times its acceleration to calculate the X component off the force after we have to that compose it as follows. It has a Y component which points in this direction, and then it's X component. These is after component why, and this is after components X. As you can see, this angle is 90 degrees. If these is 52 degrees, then this order angle here must be an angle off. 19 miners, 52 at 19 minus 52 is 38 degrees and then using this triangle. We can get a relation between the magnitude off after two on its sex component, So using the sign, we get the following The sign off 38 degrees is equal to the opposite side after you. Component acts divided by the have continues after then after Component X is given by F two times the sign off 38. Then, using that result in the expression for the second law that we have, we get that the acceleration the X direction is given by F one plus F troop times that sign off 38 degrees divided by the mass off the duck. They're plug in the values that the problem gives us. We got a X equals truth 0.1 plus 0.2 times in a sign off 38 degrees, divided by the mass off 2.5 kilograms. These results in an axe acceleration off approximately zero 0.89 true five meters per second squared. And now that they have the acts acceleration, we proceed to calculate the Y acceleration. For that, I have to organize my board. So give me a second. Okay, so for the white component noodle second low tells us that the net force is given by the mass times y acceleration in the Y direction. There is only one for acting, which is the y component off the force of truth. Then we have minus F two component white miners because actually component white points to the negative y direction is equals to the mass off the dark times. It's why acceleration, then the ducks. Why acceleration is minus after component? Why divided by the mass? Now looking at this triangle again, we can use now the co sign off 38 to get the relation between F two component way on F two. So three co sign off 38 degrees. He's given by the address inside as to component. Why divided by the magnitude after two, which is equivalent to the high Ponta news of the triangle. Then, after you component wine is equals to have to time the co sign off 38 degrees, said the why acceleration is minus F two times the curl sign off 38 degrees, divided by the mass the duck. Then plug in the values that the problem give us. We have minus 0.2 times the co sign off 38 degrees divided by troop on five and these results in an acceleration off approximately miners. Zero 0.63 04 meters per second squared. And now that we have brief accelerations, we can complete the displacement off the duck for that. Let me organize my board. Okay, So the ex displacement for a look is given by the following equation. Displacement is equals to the initial velocity times the interval of time plus the acceleration. The X direction times the interval of time squared, divided by truth. Then it's X displacement is given by 0.11 because the velocity points to the X direction times treat seconds plus 0.89 true, five times three squared. Divided by truth. These results in its placement there is equals to zero 0.73 16 meters in the X direction for the Y direction we have. The wide displacement is given by sin. Initial velocity in the Y direction times fainter off time was it's why acceleration times that interval of time squared divided by two its initial velocity in the Y direction is it close to zero because it's not moving in that direction at all. V wants to the right. Only then it's why displacement is given by minus 0.6304 times three squared, divided by truth and these results in a displacement off approximately minus 0.2837 meters in the wider action. Then we can calculate its total displacement using a triangle to it had a displacement. The X direction. It actually has a displacement in the Y direction. They make a 90 degree angle disease. The Y displacement is the axe displacement, and these is the Nat Displacement. Then, using the Pythagorean fury, um, you can calculate the net displacement as the square it off the Weitz placements Weird plus the X displacement squared. And these gives the square it off 0.7316 squared plus 0.2837 squared minus. Sine squared is just a plus sign. And, yes, results in the net displacement off approximately zero 0.78 meters. And then for the direction off the displacement, we can do the following. Let us say that this angle is Tita. Then the tangent off Peter is given by the airport excites. Divided by the address in sight, The reform teeter is the inverse tangent, off zero 0.7316 divided by zero point True, 837 noticed that I'm not including the minor sign, because in the situation, it's just one of the sides of a triangle on the side of a triangle is always positive. Then we conclude that teeter is approximately 68 0.8 degrees. So this is the east direction. Now we can calculate this angle that the displacement makes with the east directions. Let me call it Alfa, so we know that Alpha plus Tita should be 90 degrees. But Tita is 68.8. Then the angle between the displacement and the X direction Alfa is given by 90 minus 68 point in the race, and these results in angle off approximately 21 degrees, then the direction off displacement is 21 degrees south off east undies is down. Sort of this question