Problem 24 - 5 Steps to A 5: AP Calculus AB 2021

Question

The velocity function of a particle moving along the \(x\) -axis is \(v(t)=t\) \(\cos \left(t^{2}+1\right)\) for \(t \geq 0\) (a) If at \(t=0,\) the particle is at the origin, find the position of the particle at \(t=2\). (b) Is the particle moving to the right or left at \(t=2 ?\) (c) Find the acceleration of the particle at \(t=2\) and determine if the velocity of the particle is increasing or decreasing. Explain why.

Step-by-Step Solution

Answer

#Question#: At t=2, determine the position of a particle moving in a straight line with a velocity $v(t) = t\cos(t^2 + 1)$. Find the particle's direction at t=2 and explain if the velocity is increasing or decreasing at this time. #Answer#: At t=2, the position of the particle is approximately -1.3018. The particle is moving to the left, as the velocity at t=2 is negative. The velocity is increasing at t=2 because the acceleration and velocity have opposite signs, meaning that the particle is decelerating as it moves to the left.

Step 1: Integrate the velocity function to obtain the position function

To find the position function, we need to integrate the velocity function: $$ x(t) = \int v(t) dt = \int t\cos(t^2 + 1) dt $$

Step 2: Evaluate the position function at t=0

To find the constant of integration, evaluate the position function at t=0, knowing that the particle is at the origin (x=0): $$ 0 = \int_0^0 t\cos(t^2 + 1) dt \implies C = 0 $$ Thus, the position function is: $$ x(t) = \int t\cos(t^2 + 1) dt $$

Step 3: Find the position of the particle at t=2

Now we need to evaluate the position function at t=2 to find the position of the particle: $$ x(2) = \int_0^2 2\cos(2^2 + 1) dt \approx -1.3018 $$ Therefore, the position of the particle at t=2 is approximately -1.3018.

Step 4: Evaluate the velocity function at t=2

To determine the direction of motion of the particle, we need to evaluate the velocity function at t=2: $$ v(2) = 2\cos(2^2 + 1) \approx -2.4161 $$ The particle is moving to the left at t=2, since v(2) < 0.

Step 5: Differentiate the velocity function to obtain the acceleration function

To find the acceleration function, we need to differentiate the velocity function: $$ a(t) = \frac{d}{dt}\left(t\cos(t^2 + 1)\right) = \cos(t^2 + 1) - 2t^2\sin(t^2 + 1) $$

Step 6: Find the acceleration at t=2

Now we need to evaluate the acceleration function at t=2 to find its value: $$ a(2) = \cos(2^2 + 1) - 2(2^2)\sin(2^2 + 1) \approx 13.162 $$

Step 7: Determine if the velocity is increasing or decreasing

Since we have found the acceleration at t=2 (a(2) > 0) and the velocity at t=2 is negative (v(2) < 0), we can conclude that the velocity is increasing at t=2 because the acceleration and velocity have opposite signs. The particle is decelerating as it moves to the left.

Key Concepts

Velocity Function

A velocity function describes how the speed and direction of a particle change over time. In this problem, the velocity function is given by \(v(t) = t \cos(t^2 + 1)\) for \(t \geq 0\). This function tells us everything about the motion of the particle, allowing us to determine its speed and direction at any point in time by plugging various values of \(t\) into it.

• \(v(t) > 0\): The particle moves right.
• \(v(t) < 0\): The particle moves left.
• \(v(t) = 0\): The particle is at rest.

By evaluating the velocity at \(t = 2\), we found \(v(2) = -2.4161\), which shows that the particle is moving to the left at that instant.

Position Function

The position function provides insight into the exact location of a particle at any given time. It is derived from the velocity function by integration. If we know how far the object travels over time, we can calculate where it ends up. This is represented by \(x(t) = \int v(t) \, dt = \int t \cos(t^2 + 1) \, dt\).Given the condition that the particle is at the origin \((x = 0)\) at \(t = 0\), we conclude that the constant of integration is \(0\). This gives us \(x(t) = \int t \cos(t^2 + 1) \, dt\).After evaluating this integral from \(t = 0\) to \(t = 2\), the position at \(t = 2\) is approximately \(-1.3018\). This represents the particle's location along the \(x\)-axis.

Acceleration

Acceleration is the rate at which the velocity of a particle changes over time. It gives us an idea of how quickly the object is speeding up or slowing down. The acceleration function is found by differentiating the velocity function: \(a(t) = \frac{d}{dt}\left(t \cos(t^2 + 1)\right) = \cos(t^2 + 1) - 2t^2 \sin(t^2 + 1)\).By calculating the acceleration at \(t = 2\), \(a(2) \approx 13.162\), we see the particle experiences positive acceleration.

Integration

Integration is a powerful tool in calculus used to find areas under curves, as well as to derive functions such as the position given velocity. In the context of calculating the position from the velocity function, we perform the operation:\[ x(t) = \int v(t) \, dt = \int t \cos(t^2 + 1) \, dt \]Integration gives us a family of functions, represented by \(C\), the constant of integration. Using initial conditions, such as the particle being at the origin at time \(t = 0\), we can solve for \(C\), ensuring our position function accurately represents motion.

Differentiation

Differentiation is the process of finding a derivative, which represents how one quantity changes with respect to another. It's essential in finding rates of change such as velocity and acceleration. To find the acceleration function from the velocity function:\[ a(t) = \frac{d}{dt}(t \cos(t^2 + 1)) = \cos(t^2 + 1) - 2t^2 \sin(t^2 + 1) \]Differentiation simplifies complex situations, helping us identify how variables interact dynamically over time. In this scenario, we used differentiation to explore the particle's changing velocity, uncovering its acceleration characteristics at specific points.