The path r(t) = (t) + (21²-5) j describes motion on the parabola y = 2x2-5. Find the particle's velocity and acceleration vectors at t= 1, and sketch them as vectors on the curve. The velocity vector at t= 1 is v(1)=i+j (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)
The path r(t) = (t) + (21²-5) j describes motion on the parabola y = 2x2-5. Find the particle's velocity and acceleration vectors at t= 1, and sketch them as vectors on the curve. The velocity vector at t= 1 is v(1)=i+j (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)
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![**Vector Functions and Motion Along a Curve**
The path **r(t) = ⟨t, (2t² - 5)⟩** describes motion on the parabola **y = 2x² - 5**. Find the particle's velocity and acceleration vectors at **t = 1**, and sketch them as vectors on the curve.
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**The velocity vector at t = 1 is v(1) = ⟨[ ], [ ]⟩** *(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)*
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Explanation:
1. **Velocity Vector**: To find the velocity vector, differentiate the position vector **r(t)** with respect to time **t**.
If **r(t) = ⟨t, (2t² - 5)⟩**, then:
- **v(t)** = d**r(t)**/dt = ⟨1, 4t⟩
So, at t = 1:
- **v(1)** = ⟨1, 4⟩
2. **Acceleration Vector**: To find the acceleration vector, differentiate the velocity vector **v(t)** with respect to time **t**.
If **v(t) = ⟨1, 4t⟩**, then:
- **a(t)** = d**v(t)**/dt = ⟨0, 4⟩
So, at t = 1, the acceleration vector is:
- **a(1)** = ⟨0, 4⟩
3. **Sketching Vectors**: When sketching the vectors on the curve, make sure to:
- Draw the parabola **y = 2x² - 5**.
- Plot the point corresponding to **t = 1**, which is **(1, -3)**.
- Draw the velocity vector **v(1) = ⟨1, 4⟩** originating from the point **(1, -3)**.
- Draw the acceleration vector **a(1) = ⟨0, 4⟩** also originating from the point **(1, -3)**.
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By following these steps, you will be able to find and sketch the velocity](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F41a1af84-f79e-495e-9bf6-e155c46aa5b8%2F431ccff5-dfa9-4f71-a6e5-c26a1fc04402%2F84pjgke_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Vector Functions and Motion Along a Curve**
The path **r(t) = ⟨t, (2t² - 5)⟩** describes motion on the parabola **y = 2x² - 5**. Find the particle's velocity and acceleration vectors at **t = 1**, and sketch them as vectors on the curve.
---
**The velocity vector at t = 1 is v(1) = ⟨[ ], [ ]⟩** *(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)*
---
Explanation:
1. **Velocity Vector**: To find the velocity vector, differentiate the position vector **r(t)** with respect to time **t**.
If **r(t) = ⟨t, (2t² - 5)⟩**, then:
- **v(t)** = d**r(t)**/dt = ⟨1, 4t⟩
So, at t = 1:
- **v(1)** = ⟨1, 4⟩
2. **Acceleration Vector**: To find the acceleration vector, differentiate the velocity vector **v(t)** with respect to time **t**.
If **v(t) = ⟨1, 4t⟩**, then:
- **a(t)** = d**v(t)**/dt = ⟨0, 4⟩
So, at t = 1, the acceleration vector is:
- **a(1)** = ⟨0, 4⟩
3. **Sketching Vectors**: When sketching the vectors on the curve, make sure to:
- Draw the parabola **y = 2x² - 5**.
- Plot the point corresponding to **t = 1**, which is **(1, -3)**.
- Draw the velocity vector **v(1) = ⟨1, 4⟩** originating from the point **(1, -3)**.
- Draw the acceleration vector **a(1) = ⟨0, 4⟩** also originating from the point **(1, -3)**.
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By following these steps, you will be able to find and sketch the velocity
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