Question
A car travels due east at 22 m/s. It makes a turn due south and continues to travel at 22m/s. What is the change in velocity of the car?
Answers
GPT-4o mini
To find the change in velocity of the car, we need to consider its initial and final velocity vectors.
1. **Initial Velocity (\( \vec{v}_i \))**: The car is initially traveling due east at a speed of 22 m/s. We can represent this velocity as a vector:
\[
\vec{v}_i = 22 \, \text{m/s} \, \hat{i}
\]
(where \( \hat{i} \) represents the east direction).
2. **Final Velocity (\( \vec{v}_f \))**: After turning due south, the car continues to travel at the same speed of 22 m/s. We can represent this final velocity as:
\[
\vec{v}_f = 22 \, \text{m/s} \, \hat{j}
\]
(where \( \hat{j} \) represents the south direction).
3. **Change in Velocity (\( \Delta \vec{v} \))**: The change in velocity is given by the final velocity minus the initial velocity:
\[
\Delta \vec{v} = \vec{v}_f - \vec{v}_i
\]
Substituting the vectors:
\[
\Delta \vec{v} = 22 \, \text{m/s} \, \hat{j} - 22 \, \text{m/s} \, \hat{i}
\]
This can be written as:
\[
\Delta \vec{v} = -22 \, \text{m/s} \, \hat{i} + 22 \, \text{m/s} \, \hat{j}
\]
4. **Magnitude of the Change in Velocity**: To find the magnitude of this change in velocity vector, we can use the Pythagorean theorem:
\[
|\Delta \vec{v}| = \sqrt{(-22)^2 + (22)^2}
\]
\[
|\Delta \vec{v}| = \sqrt{484 + 484} = \sqrt{968} \approx 31.1 \, \text{m/s}
\]
5. **Direction of the Change in Velocity**: The direction can be calculated using the arctangent of the components:
\[
\theta = \tan^{-1}\left(\frac{22 \, \text{m/s}}{-22 \, \text{m/s}}\right) = \tan^{-1}(-1) \approx -45^\circ
\]
Since the change in velocity vector points from the initial velocity (east) to the final velocity (south), the angle corresponds to a vector heading south-west.
Thus, the result for the change in velocity of the car is approximately:
\[
|\Delta \vec{v}| \approx 31.1 \, \text{m/s}
\] in a direction that can be described as south-west.
1. **Initial Velocity (\( \vec{v}_i \))**: The car is initially traveling due east at a speed of 22 m/s. We can represent this velocity as a vector:
\[
\vec{v}_i = 22 \, \text{m/s} \, \hat{i}
\]
(where \( \hat{i} \) represents the east direction).
2. **Final Velocity (\( \vec{v}_f \))**: After turning due south, the car continues to travel at the same speed of 22 m/s. We can represent this final velocity as:
\[
\vec{v}_f = 22 \, \text{m/s} \, \hat{j}
\]
(where \( \hat{j} \) represents the south direction).
3. **Change in Velocity (\( \Delta \vec{v} \))**: The change in velocity is given by the final velocity minus the initial velocity:
\[
\Delta \vec{v} = \vec{v}_f - \vec{v}_i
\]
Substituting the vectors:
\[
\Delta \vec{v} = 22 \, \text{m/s} \, \hat{j} - 22 \, \text{m/s} \, \hat{i}
\]
This can be written as:
\[
\Delta \vec{v} = -22 \, \text{m/s} \, \hat{i} + 22 \, \text{m/s} \, \hat{j}
\]
4. **Magnitude of the Change in Velocity**: To find the magnitude of this change in velocity vector, we can use the Pythagorean theorem:
\[
|\Delta \vec{v}| = \sqrt{(-22)^2 + (22)^2}
\]
\[
|\Delta \vec{v}| = \sqrt{484 + 484} = \sqrt{968} \approx 31.1 \, \text{m/s}
\]
5. **Direction of the Change in Velocity**: The direction can be calculated using the arctangent of the components:
\[
\theta = \tan^{-1}\left(\frac{22 \, \text{m/s}}{-22 \, \text{m/s}}\right) = \tan^{-1}(-1) \approx -45^\circ
\]
Since the change in velocity vector points from the initial velocity (east) to the final velocity (south), the angle corresponds to a vector heading south-west.
Thus, the result for the change in velocity of the car is approximately:
\[
|\Delta \vec{v}| \approx 31.1 \, \text{m/s}
\] in a direction that can be described as south-west.