Math  /  Geometry

Questionrunner's total displacement? dirane travel 2.5 km at an angle of 3535^{\circ} to the ground, then changes the mand travels 5.2 km at an angle of 2222^{\circ} to the ground. What is
3. A

A tiger trainer in a zoo runs 8.0 m north, turns 3535^{\circ} east of north, and runs 3.5 m . Then, after waiting for the tiger to come near, the trainer turns due east and runs 5.0 m to get the tiger in the cage. What is the trainer's total displacement?

Studdy Solution

STEP 1

1. Displacement is a vector quantity that has both magnitude and direction.
2. We will use trigonometry to resolve each segment of the trainer's path into its north-south and east-west components.
3. The angles given are measured from the north direction.

STEP 2

1. Resolve the first segment of the path into components.
2. Resolve the second segment of the path into components.
3. Resolve the third segment of the path into components.
4. Sum all the components to find the total displacement vector.
5. Calculate the magnitude and direction of the total displacement.

STEP 3

The first segment of the path is 8.0 m directly north. This means: - North component: 8.0m 8.0 \, \text{m} - East component: 0.0m 0.0 \, \text{m}

STEP 4

The second segment of the path is 3.5 m at 35 35^\circ east of north. To resolve this into components: - North component: 3.5cos(35) 3.5 \cos(35^\circ) - East component: 3.5sin(35) 3.5 \sin(35^\circ)

STEP 5

The third segment of the path is 5.0 m due east. This means: - North component: 0.0m 0.0 \, \text{m} - East component: 5.0m 5.0 \, \text{m}

STEP 6

Sum all the north components and all the east components to find the total displacement vector: - Total north component: 8.0+3.5cos(35)+0.0 8.0 + 3.5 \cos(35^\circ) + 0.0 - Total east component: 0.0+3.5sin(35)+5.0 0.0 + 3.5 \sin(35^\circ) + 5.0

STEP 7

Calculate the numerical values: - Total north component: 8.0+3.5×0.819210.8672m 8.0 + 3.5 \times 0.8192 \approx 10.8672 \, \text{m} - Total east component: 3.5×0.5736+5.07.0076m 3.5 \times 0.5736 + 5.0 \approx 7.0076 \, \text{m}

STEP 8

Calculate the magnitude of the total displacement using the Pythagorean theorem: Magnitude=(10.8672)2+(7.0076)2 \text{Magnitude} = \sqrt{(10.8672)^2 + (7.0076)^2}

STEP 9

Calculate the direction of the total displacement using the tangent function: Direction=tan1(7.007610.8672) \text{Direction} = \tan^{-1}\left(\frac{7.0076}{10.8672}\right)

STEP 10

Calculate the numerical values: - Magnitude: (10.8672)2+(7.0076)212.87m \sqrt{(10.8672)^2 + (7.0076)^2} \approx 12.87 \, \text{m} - Direction: tan1(7.007610.8672)32.5 \tan^{-1}\left(\frac{7.0076}{10.8672}\right) \approx 32.5^\circ east of north
The trainer's total displacement is approximately 12.87m 12.87 \, \text{m} at an angle of 32.5 32.5^\circ east of north.

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