Question

The velocity of a moving object, for \(t \geq 0,\) is \(\mathbf{r}^{\prime}(t)=\langle 60,96-32 t\rangle \mathrm{ft} / \mathrm{s}\) a. When is the vertical component of velocity of the object equal to \(0 ?\) b. Find \(r(t)\) if \(r(0)=(0,3)\).

Short answer

Answer: The vertical component of velocity is equal to 0 at t = 3 seconds. The position function r(t) is given by r(t) = 〈60t, 96t - 16t^2 + 3〉.

Step-by-step solution

Identify the vertical component of the velocity function

The velocity function is given by \(\mathbf{r}'(t) = \langle 60, 96-32t\rangle\). The second element of this vector, \(96 - 32t\), represents the vertical component of velocity.

Set the vertical component to \(0\) and solve for \(t\)

To find when the vertical component of velocity is \(0\), we set the vertical component, \(96-32t\), to \(0\) and solve for \(t\): $$96 - 32t = 0$$ $$t = \frac{96}{32}$$ $$t = 3$$ The vertical component of velocity is equal to \(0\) at \(t=3\) seconds. #b. Find \(r(t)\) if \(r(0) = (0,3)\)

Integrate the velocity function component by component

To find the position function, we need to integrate the velocity function component by component. So, we'll integrate the horizontal and vertical components separately. $$r_x(t) = \int 60 dt = 60t + C_x$$ $$r_y(t) = \int (96 - 32t) dt = 96t - 16t^2 + C_y$$

Apply the initial conditions to find \(C_x\) and \(C_y\)

We're given that \(r(0) = (0, 3)\). This means that when \(t=0\), \(r_x(0) = 0\) and \(r_y(0) = 3\). We can use this information to find the constants \(C_x\) and \(C_y\). $$r_x(0) = 60 \cdot 0 + C_x = 0 \Rightarrow C_x = 0$$ $$r_y(0) = 96 \cdot 0 - 16 \cdot 0^2 + C_y = 3 \Rightarrow C_y = 3$$ Now we can write the position function \(r(t)\) with its components: $$r(t) = \langle 60t, 96t - 16t^2 + 3\rangle$$

Key concepts

Vertical Component

The vertical component in a velocity context refers to the part of the velocity vector pointing upwards or downwards. In our exercise, the velocity of an object is given by the vector \( \mathbf{r}'(t) = \langle 60, 96-32t \rangle \). Here, \( 96-32t \) represents the vertical component. This component changes with time because of the inclusion of the \(-32t \) term.
This indicates the effect of gravity or any other downward force over time.

• Understanding the vertical component is essential because it affects how high or low the object will move over time.
• Setting the vertical component to zero allows us to find when the object reaches its peak point before coming back down.
• In our case, solving \( 96 - 32t = 0 \) gives \( t = 3 \). At 3 seconds, the object's upward velocity becomes zero, marking the highest point before it begins to descend.

Position Function

In physics, the position function describes the location of an object at any given time. It's derived by integrating the velocity function over time. The exercise provides the velocity vector \( \mathbf{r}'(t) = \langle 60, 96-32t \rangle \), which has both horizontal and vertical components.
To find the position function \( r(t) \), we integrate each component separately. Consider the following points:

• Integrating the horizontal component \( 60 \) gives the position function for the horizontal motion: \( r_x(t) = 60t + C_x \).
• The vertical component \( 96 - 32t \) integrates to give \( r_y(t) = 96t - 16t^2 + C_y \).
• These functions together form the overall position function \( r(t) = \langle 60t, 96t - 16t^2 + C_y \rangle \).

Integrating velocity components helps track an object's movement path over time, pinned by initial conditions.

Initial Conditions

Initial conditions are pivotal in determining the unique trajectory and position of an object over time. They refer to the known facts about an object's position or movement at the starting point, usually when \( t = 0 \). In our exercise, the object starts at \( r(0) = (0, 3) \).

• This means that at \( t = 0 \), the object is positioned 3 units up from the origin while horizontally at the origin.
• These conditions allow us to solve for any integration constants, making our position function accurate and applicable to real-world situations.

For our scenario:

• In the horizontal direction, the initial condition leads to \( C_x = 0 \), since \( 60 \cdot 0 + C_x = 0 \).
• In the vertical direction, the initial position gives us \( C_y = 3 \), because \( 96 \cdot 0 - 16 \cdot 0^2 + C_y = 3 \).

By incorporating these conditions, we tailored our position function to fit the specific path and timing of the object, guaranteeing that our calculations align with reality. Initial conditions are essential for solving differential equations in real-world applications.