Question

A grasshopper makes four jumps. The displacement vectors are (1) 31.0 cm, due west; (2) 26.0 cm, 44.0 ° south of west; (3) 22.0 cm, 56.0 ° south of east; and (4) 23.0 cm, 75.0 ° north of east. Find (a) the magnitude and (b) direction of the resultant displacement. Express the direction as a positive angle with respect to due west.

Answer 1

To find the resultant displacement of the grasshopper, we can break down the vectors into their x and y components, and then sum up the components separately. After performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction is approximately 38.3° south of west.

Step-by-step explanation:

To find the resultant displacement of the grasshopper, we need to add the individual displacement vectors. We can do this by breaking down each vector into its x and y components.

For vector (1) with a magnitude of 31.0 cm due west, the x component is -31.0 cm and the y component is 0.

Similarly, for the other vectors, the x and y components are:

• (2): x = -26.0*cos(44.0) cm, y = -26.0*sin(44.0) cm
• (3): x = 22.0*cos(56.0) cm, y = -22.0*sin(56.0) cm
• (4): x = 23.0*cos(75.0) cm, y = 23.0*sin(75.0) cm

Now, we can sum up the x components and y components separately to find the resultant displacement.

The magnitude of the resultant displacement can be found using the formula:

resultant magnitude = sqrt((sum of x components)^2 + (sum of y components)^2)

The direction of the resultant displacement can be found using the formula:

resultant direction = atan2((sum of y components), (sum of x components))

Plugging in the values and performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction of the resultant displacement is approximately 38.3° south of west.

Answer 2

(a) 34.47 cm

(b)
`24.09^\circ` south of west

Step-by-step explanation:

Let us draw a figure representing the individual displacement vectors in the four jumps as shown in the figure attached with this solution.

Now, let us try to write the four displacement vectors in in terms of unit vectors along the horizontal and the vertical axis.

`\vec{d}_1= 31\ cm\ west = -31\ cm\ \hat{i}\\\vec{d}_2= 26\ cm\ south\ of\ west = -26\cos 44^\circ\ \hat{i} -26 \sin 44^\circ\ \hat{j}=(-18.72\ \hat{i}-18.06\ \hat{i})\ cm\\\vec{d}_3= 22\ cm\ south\ of\ east = 22\cos 56^\circ\ \hat{i} -22 \sin 56^\circ\ \hat{j}=(12.30\ \hat{i}-18.23\ \hat{i})\ cm\\\vec{d}_4= 23\ cm\ north\ of\ east = 23\cos 75^\circ\ \hat{i} +23\sin \sin 75^\circ\ \hat{j}=(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\`

Now, the vector sum of all these vector will give the resultant displacement vector.

`\vec{D} = \vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} = -31\ cm\ \hat{i}+(-18.72\ \hat{i}-18.06\ \hat{i})\ cm+(12.30\ \hat{i}-18.23\ \hat{i})\ cm+(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\\Rightarrow \vec{D} =(-31.47\ \hat{i}-14.07\ \hat{i})\ cm`

Part (a):

The magnitude of the resultant displacement vector is given by:

`D=√((-31.47)^2+(-14.07)^2)\ m = 34.47\ m`

Part (b):

Since the resultant displacement vector indicates that the final position of the vector lies in the third quadrant, the vector will make some positive angle in the direction south of west given by:

`\theta = \tan^(-1)((14.07)/(31.47))= 24.09^\circ`