Question

Solving equations of motion Given an acceleration vector, initial velocity \(\left\langle u_{0}, v_{0}, w_{0}\right\rangle,\) and initial position \(\left\langle x_{0}, y_{0}, z_{0}\right\rangle,\) find the velocity and position vectors, for \(t \geq \mathbf{0}\). $$\mathbf{a}(t)=\left\langle t, e^{-t}, 1\right\rangle,\left\langle u_{0^{\prime}}, v_{0^{\prime}} w_{0}\right\rangle=\langle 0,0,1\rangle, \left\langle x_{0}, y_{0}, z_{0}\right\rangle=\langle 4,0,0\rangle$$

Short answer

Based on the given information, the position vector is: $$ \mathbf{r}(t)=\left\langle \frac{t^3}{6} + 4 , -t - e^{-t} , \frac{t^2}{2} + t \right\rangle $$

Step-by-step solution

Find the velocity vector

To find the velocity vector, we need to integrate the acceleration vector components with respect to time: $$\mathbf{v}(t)=\int{\mathbf{a}(t)}\,dt$$ Integrate each component of the acceleration vector separately: $$ v_x(t)=\int{t}\,dt\\ v_y(t)=\int{e^{-t}}\,dt\\ v_z(t)=\int{1}\,dt $$ Then add the initial velocities: $$ v_x(t)=\int{t}\,dt + u_0\\ v_y(t)=\int{(-e^{-t})}\,dt + v_0'\\ v_z(t)=\int{1}\,dt + w_0 $$

Calculate the velocity vector

Now, perform the integration for each component and add the initial velocities: $$ v_x(t)=\frac{t^2}{2} + 0\\ v_y(t)=e^{-t} + c_1 - e^{-t} + 0\\ v_z(t)=t + 1 $$ where c_1 is the constant of integration for \(v_y(t)\).

Find the position vector

To find the position vector, we need to integrate the velocity vector components with respect to time: $$\mathbf{r}(t)=\int{\mathbf{v}(t)}\,dt$$ Integrate each component of the velocity vector separately: $$ x(t)=\int{\frac{t^2}{2}}\,dt\\ y(t)=\int{(-1 + e^{-t})}\,dt\\ z(t)=\int{(t+1)}\,dt $$ Then add the initial positions: $$ x(t)=\int{\frac{t^2}{2}}\,dt + x_0\\ y(t)=\int{(-1 + e^{-t})}\,dt + y_0\\ z(t)=\int{(t+1)}\,dt + z_0 $$

Calculate the position vector

Now, perform the integration for each component and add the initial positions: $$ x(t)=\frac{t^3}{6} + 4\\ y(t)=-t + c_2 - e^{-t}\\ z(t)=\frac{t^2}{2} + t + z_0 $$ where c_2 is the constant of integration for \(y(t)\). So, the final position vector is: $$ \mathbf{r}(t)=\left\langle \frac{t^3}{6} + 4 , -t - e^{-t} , \frac{t^2}{2} + t \right\rangle $$

Key concepts

Acceleration Vector

An acceleration vector provides us with information on how the velocity of an object changes over time. In physics, particularly in kinematics, acceleration is defined as the rate at which an object's velocity changes with respect to time.

The acceleration vector in our exercise is given by the function \(\mathbf{a}(t) = \left\langle t, e^{-t}, 1\right\rangle\), which means the acceleration in the x-direction varies linearly with time (\(t\)), in the y-direction it changes as an exponential decay (\(e^{-t}\)), and in the z-direction it remains constant. To visualize this, imagine a car accelerating differently in each direction; that's what the components of the acceleration vector represent.

Integrating this vector leads to the velocity vector, as the change in velocity over time is acceleration. Therefore, integrating acceleration with respect to time gives the total change in velocity from the initial value, provided the body starts from rest or with an initial velocity vector.

Velocity Vector Integration

Velocity vector integration is the process of finding the velocity of an object at any time based on its acceleration. By integrating the acceleration vector with respect to time, we obtain a velocity vector that shows how fast and in what direction the object is moving.

In the solution, the integrals \(v_x(t) = \int{t}\,dt\), \(v_y(t) = \int{e^{-t}}\,dt\), and \(v_z(t) = \int{1}\,dt\) are calculated, with initial conditions factored in as well: \(\mathbf{v}(t) = \left\langle \frac{t^2}{2}, -e^{-t}, t + 1 \right\rangle\). This gives us a concrete expression for the velocity vector at any time \(t\).

The process of integration adds constants of integration, which are resolved by using initial conditions. For example, the given initial velocity \(\left\langle u'_0, v'_0, w_0 \right\rangle = \langle 0,0,1\rangle\) assures that the constants don't alter the initial state of the velocity.

Position Vector Calculus

Position vector calculus involves finding the location of an object at any given time when its velocity vector is known. Just as velocity is the integral of acceleration, position is the integral of velocity.

The solution to our problem required integrating the velocity function to find the position function: \(\mathbf{r}(t) = \int{\mathbf{v}(t)}\,dt\). The integrals \(x(t)=\int{\frac{t^2}{2}}\,dt\), \(y(t)=\int{(-1 + e^{-t})}\,dt\), and \(z(t)=\int{(t+1)}\,dt\) combine the integrated velocity components with the initial position \(\left\langle x_0, y_0, z_0\right\rangle = \langle 4,0,0\rangle\) to provide the object's position at any time.

Finally, by adding the initial position to the integrated velocity vector, we get \(\mathbf{r}(t)=\left\langle \frac{t^3}{6} + 4 , -t - e^{-t} , \frac{t^2}{2} + t \right\rangle\), completing the description of the object's motion. This shows us precisely where the object will be at any time \(t\), creating a full trajectory based on its motion.