From everyday questions to specialized queries, IDNLearn.com has the answers. Ask anything and receive comprehensive, well-informed responses from our dedicated team of experts.
Sagot :
To solve the problem of finding the change in velocity of the car moving east and then north, let's break down the steps in detail:
### Step-by-Step Solution:
1. Determine the Initial Velocities:
- The car moves with a constant speed of 40 m/s towards the east.
- The car then moves with a constant speed of 30 m/s towards the north.
2. Resultant Speed Calculation:
- To find the net change in velocity (resultant velocity), we can treat the eastward and northward speeds as the legs of a right triangle. We then use the Pythagorean theorem to determine the magnitude of the resultant velocity.
- [tex]\( \text{Resultant Speed} = \sqrt{(40\, \text{m/s})^2 + (30\, \text{m/s})^2} \)[/tex]
3. Substituting the given values:
- [tex]\( \text{Resultant Speed} = \sqrt{1600 + 900} \)[/tex]
- [tex]\( \text{Resultant Speed} = \sqrt{2500} \)[/tex]
- [tex]\( \text{Resultant Speed} = 50\, \text{m/s} \)[/tex]
4. Resultant Direction Calculation:
- To find the direction of the resultant velocity, we use trigonometry. Specifically, we use the arctangent function to determine the angle relative to the east direction.
- [tex]\( \theta = \tan^{-1} \left( \frac{\text{Speed North}}{\text{Speed East}} \right) \)[/tex]
- [tex]\( \theta = \tan^{-1} \left( \frac{30}{40} \right) \)[/tex]
5. Calculating the angle:
- [tex]\( \theta = \tan^{-1} (0.75) \)[/tex]
- [tex]\( \theta \approx 36.87^\circ \)[/tex]
6. Directional Adjustment:
- The angle of [tex]\( 36.87^\circ \)[/tex] is measured from the east direction towards the north.
- To determine the direction with respect to the reference options provided, we need to convert this angle to a standard bearing which is typically measured clockwise from the north.
- Since [tex]\( 36.87^\circ \)[/tex] is from the east towards north, it implies that measured counterclockwise from north, it would be [tex]\( 90^\circ - 36.87^\circ \)[/tex]
7. Converting to Bearing:
- [tex]\( 90^\circ - 36.87^\circ = 53.13^\circ \)[/tex] (north of east)
- To convert this to a bearing which we measure clockwise from north:
- [tex]\( 360^\circ - 53.13^\circ = 306.87^\circ \)[/tex]
### Conclusion:
The resultant change in velocity of the car is:
Magnitude: 50 m/s
Direction: 307° (approximately)
Thus, the correct answer is:
D 50 m/s at 307°
### Step-by-Step Solution:
1. Determine the Initial Velocities:
- The car moves with a constant speed of 40 m/s towards the east.
- The car then moves with a constant speed of 30 m/s towards the north.
2. Resultant Speed Calculation:
- To find the net change in velocity (resultant velocity), we can treat the eastward and northward speeds as the legs of a right triangle. We then use the Pythagorean theorem to determine the magnitude of the resultant velocity.
- [tex]\( \text{Resultant Speed} = \sqrt{(40\, \text{m/s})^2 + (30\, \text{m/s})^2} \)[/tex]
3. Substituting the given values:
- [tex]\( \text{Resultant Speed} = \sqrt{1600 + 900} \)[/tex]
- [tex]\( \text{Resultant Speed} = \sqrt{2500} \)[/tex]
- [tex]\( \text{Resultant Speed} = 50\, \text{m/s} \)[/tex]
4. Resultant Direction Calculation:
- To find the direction of the resultant velocity, we use trigonometry. Specifically, we use the arctangent function to determine the angle relative to the east direction.
- [tex]\( \theta = \tan^{-1} \left( \frac{\text{Speed North}}{\text{Speed East}} \right) \)[/tex]
- [tex]\( \theta = \tan^{-1} \left( \frac{30}{40} \right) \)[/tex]
5. Calculating the angle:
- [tex]\( \theta = \tan^{-1} (0.75) \)[/tex]
- [tex]\( \theta \approx 36.87^\circ \)[/tex]
6. Directional Adjustment:
- The angle of [tex]\( 36.87^\circ \)[/tex] is measured from the east direction towards the north.
- To determine the direction with respect to the reference options provided, we need to convert this angle to a standard bearing which is typically measured clockwise from the north.
- Since [tex]\( 36.87^\circ \)[/tex] is from the east towards north, it implies that measured counterclockwise from north, it would be [tex]\( 90^\circ - 36.87^\circ \)[/tex]
7. Converting to Bearing:
- [tex]\( 90^\circ - 36.87^\circ = 53.13^\circ \)[/tex] (north of east)
- To convert this to a bearing which we measure clockwise from north:
- [tex]\( 360^\circ - 53.13^\circ = 306.87^\circ \)[/tex]
### Conclusion:
The resultant change in velocity of the car is:
Magnitude: 50 m/s
Direction: 307° (approximately)
Thus, the correct answer is:
D 50 m/s at 307°
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.