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A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 73 ◦ with the positive x axis. The result after the two displacements is 8.7 m directed at an angle of 119◦ to the positive x axis using counterclockwise as the positive angular direction Find the angle of the second displacement (measured counterclockwise from the positive x axis). Answer in units of ◦

2 Answers

1 vote

Final answer:

To find the angle of the second displacement vector with respect to the positive x-axis, one must analyze the given displacements using vector addition and trigonometric functions, accounting for both the magnitudes and angles provided.

Step-by-step explanation:

To determine the angle of the second displacement in relation to the positive x-axis, one must use vector addition and trigonometry. The first displacement vector is given to have a magnitude of 11 m and is at 73° to the positive x-axis. The resultant vector has a magnitude of 8.7 m at an angle of 119°. To find the angle of the second displacement, we need to analyze the components of these vectors and solve using the method of vector addition.

Let's break down the given displacements into their x and y components. The first displacement vector (A) has components Ax = 11 m × cos(73°) and Ay = 11 m × sin(73°). The resultant vector (R) has components Rx = 8.7 m × cos(119°) and Ry = 8.7 m × sin(119°). We can then use these to solve for the second displacement vector's components, and subsequently its magnitude B and direction angle θB.

Using vector addition, Bx = Rx - Ax and By = Ry - Ay. Then, we can use these to solve for the magnitude B and direction angle θB of the second displacement vector. The direction angle is found using the arctan function: θB = tan⁻¹ (By / Bx). This will give us the angle of the second displacement with respect to the positive x-axis.

User Aditi
by
3.0k points
6 votes

Answer:

7.97253265907 m

Step-by-step explanation:

First displacement


\Delta x=11cos 73=3.216\ m


\Delta y=11sin 73=10.51\ m

Final displacement


\Delta x=8.7cos 119=-4.21\ m


\Delta y=8.7sin 119=7.609\ m

Second displacement


\Delta x=-4.21-3.216=-7.426\ m


\Delta y=7.609-10.51=-2.901\ m

The magnitude of the second displacement is


=√((-7.426)^2+(-2.901)^2)=7.97253265907\ m

The second displacement is 7.97253265907 m

User Brian Goldman
by
3.4k points