Question

Given an acceleration vector. initial velocity \(\left\langle u_{0}, v_{0}\right\rangle,\) and initial position \(\left\langle x_{0}, y_{0}\right\rangle,\) find the velocity and position vectors, for \(t \geq 0\). $$\mathbf{a}(t)=\langle\cos t, 2 \sin t\rangle,\left\langle u_{0}, v_{0}\right\rangle=\langle 0,1\rangle,\left\langle x_{0}, y_{0}\right\rangle=\langle 1,0\rangle$$

Short answer

Question: Find the expressions for the velocity vector and position vector of a particle in two-dimensional space given the acceleration vector \(\mathbf{a}(t) = \langle \cos t, 2\sin t \rangle\), initial velocity vector \(\mathbf{v}(0) = \langle 0, 1 \rangle\), and initial position vector \(\mathbf{r}(0) = \langle 1, 0 \rangle\). Answer: The velocity vector is given by \(\mathbf{v}(t) = \langle \sin t, -2\cos t + 3 \rangle\), and the position vector is given by \(\mathbf{r}(t) = \langle -\cos t + 2, -2\sin t + 3t \rangle\).

Step-by-step solution

Integrate the acceleration vector to find the velocity vector

We are given the acceleration vector \(\mathbf{a}(t) = \langle \cos t, 2\sin t \rangle\). To find the velocity vector, integrate the acceleration vector with respect to time: $$\mathbf{v}(t) = \int \mathbf{a}(t) dt = \int \langle \cos t, 2\sin t \rangle dt$$ Integrating component-wise, we get: $$\mathbf{v}(t) = \langle \int \cos t dt, \int 2\sin t dt \rangle$$

Evaluate the integrals and apply the initial velocity condition

Calculating the integrals, we have: $$\mathbf{v}(t) = \langle \sin t + C_1, -2 \cos t + C_2 \rangle$$ We're given the initial velocity vector \(\mathbf{v}(0) = \langle u_0, v_0 \rangle = \langle 0, 1 \rangle\). To find the constants \(C_1\) and \(C_2\), substitute \(t=0\) into the expression for \(\mathbf{v}(t)\): $$\mathbf{v}(0) = \langle \sin(0) + C_1, -2 \cos(0) + C_2 \rangle = \langle 0, 1 \rangle$$ From this, we get \(C_1 = 0\) and \(C_2 = 3\). Thus, the velocity vector is: $$\mathbf{v}(t) = \langle \sin t, -2\cos t + 3 \rangle$$

Integrate the velocity vector to find the position vector

To find the position vector, integrate the velocity vector with respect to time: $$\mathbf{r}(t) = \int \mathbf{v}(t) dt = \int \langle \sin t, -2\cos t + 3 \rangle dt$$ Integrating component-wise, we get: $$\mathbf{r}(t) = \langle \int \sin t dt, \int (-2\cos t + 3) dt \rangle$$

Evaluate the integrals and apply the initial position condition

Calculating the integrals, we have: $$\mathbf{r}(t) = \langle -\cos t + C_3, -2\sin t + 3t + C_4 \rangle$$ We're given the initial position vector \(\mathbf{r}(0) = \langle x_0, y_0 \rangle = \langle 1, 0 \rangle\). To find the constants \(C_3\) and \(C_4\), substitute \(t=0\) into the expression for \(\mathbf{r}(t)\): $$\mathbf{r}(0) = \langle -\cos(0) + C_3, -2\sin(0) + 3(0) + C_4 \rangle = \langle 1, 0 \rangle$$ From this, we get \(C_3 = 2\) and \(C_4 = 0\). Thus, the position vector is: $$\mathbf{r}(t) = \langle -\cos t + 2, -2\sin t + 3t \rangle$$ In conclusion, the velocity vector is given by \(\mathbf{v}(t) = \langle \sin t, -2\cos t + 3 \rangle\), and the position vector is given by \(\mathbf{r}(t) = \langle -\cos t + 2, -2\sin t + 3t \rangle\).

Key concepts

Understanding the Acceleration Vector

The acceleration vector represents how the velocity of an object changes over time. In vector calculus, it is typically denoted as \( \mathbf{a}(t) = \langle a_x(t), a_y(t) \rangle \). In our exercise, the given acceleration vector is \( \langle \cos t, 2\sin t \rangle \). The components \( \cos t \) and \( 2\sin t \) describe how the object speeds up or slows down in the horizontal and vertical directions, respectively.

To fully understand the acceleration vector's role, one should think of it as providing instructions on how to adjust the velocity at every moment \( t \). By integrating this vector, we obtain the velocity vector, which gives a more comprehensive view of how the object's motion evolves. We'll explain this next!

Unraveling the Velocity Vector

The velocity vector gives us the rate of change of the position of an object with time. It is often expressed as \( \mathbf{v}(t) = \langle v_x(t), v_y(t) \rangle \). Velocity includes both the speed and direction of the object's motion.

From our initial velocity \( \langle 0, 1 \rangle \) and integrating the acceleration vector, we derive the velocity vector \( \mathbf{v}(t) = \langle \sin t, -2\cos t + 3 \rangle \). This expression tells us how fast and in which direction the object is moving at any time \( t \). The component \( \sin t \) describes horizontal movement, whereas \( -2\cos t + 3 \) specifies vertical motion.

The integration process involves calculating indefinite integrals of the acceleration components, adjusting with initial conditions to find constants \( C_1 \) and \( C_2 \). This ensures our velocity vector fits the given initial condition.

Getting to the Position Vector

The position vector represents an object's location in space at any given time and is typically written as \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \). In the context of motion, it is crucial to know where the object is positioned at any instant.

To find the position vector, we integrate the velocity vector \( \mathbf{v}(t) = \langle \sin t, -2\cos t + 3 \rangle \), which results in \( \mathbf{r}(t) = \langle -\cos t + 2, -2\sin t + 3t \rangle \). The process involves integrating each component of the velocity vector and applying the initial position \( \langle 1, 0 \rangle \) to solve for constants \( C_3 \) and \( C_4 \).

Understanding this vector helps us comprehend the trajectory or path of the object over time, combining its directional changes given by the velocity vector.