A grasshopper makes four jumps

A grasshopper makes four jumps. The displacement vectors are (1) 27.0 cm, due west; (2) 23.0 cm, 35o south of west; (3) 28.0 cm, 55o south of east, and (4) 35.0 cm, 63o north of east (5) 18.0 cm, east. Find the magnitude and direction of the resultant displacement.

The correct answer and explanation is :

To find the magnitude and direction of the resultant displacement, we will break the displacement vectors into their components, sum the components in the x (east-west) and y (north-south) directions, and then use the Pythagorean theorem to find the resultant displacement vector’s magnitude and direction.

Step 1: Resolve each vector into components

For each displacement vector, we can resolve it into its components using trigonometry. For each vector, the x-component is $r \cos \theta$ and the y-component is $r \sin \theta$, where $r$ is the magnitude of the vector and $\theta$ is the angle.

1. Vector 1 (27.0 cm, due west):

• $\theta = 180^\circ$ (west direction)
• $x_1 = -27.0$ cm (negative because it’s westward)
• $y_1 = 0$ cm

1. Vector 2 (23.0 cm, 35° south of west):

• $\theta = 180^\circ + 35^\circ = 215^\circ$
• $x_2 = 23.0 \cos(215^\circ) = -23.0 \times 0.819 = -18.85$ cm
• $y_2 = 23.0 \sin(215^\circ) = -23.0 \times 0.574 = -13.21$ cm

1. Vector 3 (28.0 cm, 55° south of east):

• $\theta = 55^\circ$ (south of east)
• $x_3 = 28.0 \cos(55^\circ) = 28.0 \times 0.573 = 16.06$ cm
• $y_3 = 28.0 \sin(55^\circ) = 28.0 \times 0.819 = 22.93$ cm

1. Vector 4 (35.0 cm, 63° north of east):

• $\theta = 63^\circ$
• $x_4 = 35.0 \cos(63^\circ) = 35.0 \times 0.454 = 15.91$ cm
• $y_4 = 35.0 \sin(63^\circ) = 35.0 \times 0.891 = 31.22$ cm

1. Vector 5 (18.0 cm, east):

• $\theta = 0^\circ$
• $x_5 = 18.0$ cm
• $y_5 = 0$ cm

Step 2: Sum the components

Now we sum the x- and y-components of all vectors:

• Total x-component: $$
  x_{\text{total}} = (-27.0) + (-18.85) + 16.06 + 15.91 + 18.0 = 3.12 \, \text{cm}
  $$
• Total y-component: $$
  y_{\text{total}} = 0 + (-13.21) + 22.93 + 31.22 + 0 = 41.94 \, \text{cm}
  $$

Step 3: Find the resultant displacement magnitude

The magnitude $R$ of the resultant displacement is given by the Pythagorean theorem:

$$
R = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}
$$

$$
R = \sqrt{(3.12)^2 + (41.94)^2} = \sqrt{9.7344 + 1761.48} = \sqrt{1771.21} = 42.14 \, \text{cm}
$$

Step 4: Find the direction of the resultant displacement

The direction $\theta$ of the resultant displacement is given by:

$$
\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right)
$$

$$
\theta = \tan^{-1}\left(\frac{41.94}{3.12}\right) = \tan^{-1}(13.44) = 86.23^\circ
$$

Thus, the resultant displacement is 42.14 cm, at an angle of 86.23° north of east.

Final Answer:

• Magnitude: 42.14 cm
• Direction: 86.23° north of east