Let's go through these problems step by step.
1. Displacement of the Car
The car travels 80.0 km east and then 192 km south. We can represent these movements with vector components.
- East (x-direction): \( x_1 = 80.0 , \text{km} \)
- South (y-direction): \( y_1 = -192 , \text{km} \)
To find the resultant displacement \( R \), we can use the Pythagorean theorem: \[ R = \sqrt{x_1^2 + y_1^2} = \sqrt{(80.0)^2 + (-192)^2} \] \[ R = \sqrt{6400 + 36864} = \sqrt{43264} \approx 207.93 , \text{km} \]
To find the direction (angle \( \theta \)) with respect to the east (x-axis): \[ \theta = \tan^{-1} \left( \frac{y_1}{x_1} \right) = \tan^{-1} \left( \frac{-192}{80} \right) \approx -67.38^\circ \] This angle is measured from the positive x-axis (east) toward the negative y-axis (south).
Summary for Car
- Displacement: \( \approx 207.93 , \text{km} \)
- Direction: \( 67.4^\circ \) south of east.
2. Displacement of the Turtle
The turtle ends up at point \( (24, 10) \).
The displacement from the origin (0,0) to (24, 10) can be found using: \[ R = \sqrt{x^2 + y^2} = \sqrt{(24)^2 + (10)^2} \] \[ R = \sqrt{576 + 100} = \sqrt{676} = 26 , \text{cm} \]
To find the direction: \[ \theta = \tan^{-1}\left( \frac{y}{x} \right) = \tan^{-1}\left( \frac{10}{24} \right) \approx 22.62^\circ \]
Summary for Turtle
- Displacement: \( 26 , \text{cm} \)
- Direction: \( 22.6^\circ \) above the positive x-axis.
3. Displacement of the Bug
The bug's movements can be summarized as follows:
- East: \( 8.0 , \text{cm} \)
- South: \( 5.0 , \text{cm} \)
- West: \( 3.0 , \text{cm} \)
- North: \( 4.0 , \text{cm} \)
Net displacements:
- North-South: \( 4.0 - 5.0 = -1.0 , \text{cm} \) (1 cm south)
- East-West: \( 8.0 - 3.0 = 5.0 , \text{cm} \) (5 cm east)
(a) From point A to point B:
- North: \( -1.0 , \text{cm} \) (1 cm south)
- East: \( 5.0 , \text{cm} \) (5 cm east)
(b) Find the displacement \( R \): \[ R = \sqrt{(5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.10 , \text{cm} \]
Direction: \[ \theta = \tan^{-1} \left( \frac{-1}{5} \right) \approx -11.31^\circ \]
Summary for Bug
- North/South: \( 1.0 , \text{cm} \) south
- East/West: \( 5.0 , \text{cm} \) east
- Displacement: \( \approx 5.10 , \text{cm} \)
- Direction: \( 11.3^\circ \) south of east.
4. Scalar Components of Displacements
For each angle, we will decompose the vector into x and y components using:
\[ x = r \cos(\theta), \quad y = r \sin(\theta) \]
(a) For \( 300, \text{cm} , \text{at} , 127^\circ \): \[ x = 300 \cos(127^\circ) \approx -180.95 , \text{cm} \] \[ y = 300 \sin(127^\circ) \approx 227.76 , \text{cm} \]
(b) For \( 500, \text{cm} , \text{at} , 220^\circ \): \[ x = 500 \cos(220^\circ) \approx -469.85 , \text{cm} \] \[ y = 500 \sin(220^\circ) \approx -125.00 , \text{cm} \]
Summary for Components
(a) \( x \approx -180.95 , \text{cm}, , y \approx 227.76 , \text{cm} \) (b) \( x \approx -469.85 , \text{cm}, , y \approx -125.00 , \text{cm} \)
5. Resultant Forces
Two forces given:
- \( F_1 = 100 , \text{N} \) at \( 170^\circ \)
- \( F_2 = 100 , \text{N} \) at \( 50^\circ \)
Components: \[ F_{1x} = 100 \cos(170^\circ) = -96.19 , \text{N}, \quad F_{1y} = 100 \sin(170^\circ) = 17.36 , \text{N} \] \[ F_{2x} = 100 \cos(50^\circ) \approx 64.28 , \text{N}, \quad F_{2y} = 100 \sin(50^\circ) \approx 76.94 , \text{N} \]
Resultant components: \[ R_x = F_{1x} + F_{2x} \approx -96.19 + 64.28 \approx -31.91 , \text{N} \] \[ R_y = F_{1y} + F_{2y} \approx 17.36 + 76.94 \approx 94.30 , \text{N} \]
Magnitude: \[ R = \sqrt{R_x^2 + R_y^2} \approx \sqrt{(-31.91)^2 + (94.30)^2} \approx 98.87 , \text{N} \]
Direction: \[ \theta = \tan^{-1}\left( \frac{R_y}{R_x} \right) = \tan^{-1}\left( \frac{94.30}{-31.91} \right) \approx 112.6^\circ \]
Summary for Forces
- Resultant Force: \( \approx 98.87 , \text{N} \)
- Direction: \( 112.6^\circ \) from the positive x-axis.
6. Resultant Displacement
Starting at the origin and given the movements:
- \( (0, 60) \)
- \( (-30, 0) \)
- Calculate from angle \( 150^\circ\): \[ R_3 = 40(\cos(150^\circ), \sin(150^\circ)) \approx (-20, 20) \]
- Calculate from angle \( 240^\circ\): \[ R_4 = 50(\cos(240^\circ), \sin(240^\circ)) \approx (25(-0.5), 25(-\sqrt{3}/2)) \approx (-25, -21.65) \]
Now summing these vectors:
- \( x_{\text{total}} = 0 + (-30) + (-20) + (-25) = -75 \)
- \( y_{\text{total}} = 60 + 0 + 20 + (-21.65) \approx 58.35 \)
Resultant displacement: \[ R = \sqrt{(-75)^2 + (58.35)^2} \approx 92.93 , \text{mm} \] Direction: \[ \theta = \tan^{-1}\left( \frac{58.35}{-75} \right) \approx 143.59^\circ \]
Summary for Resultants
- Resultant Displacement: \( \approx 92.93 , \text{mm} \)
- Direction: \( 143.6^\circ \) from the positive x-axis.