Chapter 12: Problem 12
Give the position vectors of particles moving along various curves in the \(x y\) -plane. In each case, find the particle's velocity and acceleration vectors at the stated times, and sketch them as vectors on the curve. Motion on the parabola \(y=x^{2}+1\) $$\mathbf{r}(t)=t \mathbf{i}+\left(t^{2}+1\right) \mathbf{j} ; \quad t=-1,0, \text { and } 1$$
Short Answer
Step by step solution
Identify the position vector
Find the velocity vector
Calculate the velocity at t = -1, 0, and 1
Find the acceleration vector
Calculate the acceleration at t = -1, 0, and 1
Sketch the vectors on the curve
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Position Vectors
This ordered pair is essentially broken down into the x-component, \( t \mathbf{i} \), and the y-component, \( (t^2 + 1) \mathbf{j} \). Each component represents movement along the x or y axis.
- The x-component \( t \mathbf{i} \) indicates motion along the horizontal x-axis.
- The y-component \( (t^2 + 1) \mathbf{j} \) reflects the particle's position along the vertical y-axis, which is also influenced by \( t \).
Velocity Vectors
This new vector indicates not just movement but **speed** and **direction**. The i-component (\(1 \mathbf{i}\)) remains constant, showing a steady motion in the x-direction, while the j-component changes with time \( 2t \mathbf{j} \), reflecting acceleration along the y-axis.
- A constant \( 1 \mathbf{i} \) means the particle moves consistently at one unit per time along x.
- The y-component \( 2t \mathbf{j} \) suggests that vertical speed increases as time passes.
Acceleration Vectors
Key insights into acceleration include:
- \( 0 \mathbf{i} \) indicates there is no change in velocity in the x-direction; so, the motion here is uniform and unchanging horizontally.
- \( 2 \mathbf{j} \) means there is a constant acceleration in the y-direction, influencing how velocity increases vertically.