Question

Motion of a particle is described by \(\mathrm{x}=(\mathrm{t}-2)^{2}\) Find its velocity when it passes through origin. (A) \(0 \mathrm{~m} / \mathrm{s}\) (B) \(2 \mathrm{~ms}^{-1}\) (C) \(4 \mathrm{~ms}^{-1}\) (D) \(8 \mathrm{~ms}^{-1}\)

Short answer

The velocity of the particle when it passes through the origin is \(0 \mathrm{~m} / \mathrm{s}\), so the answer is (A).

Step-by-step solution

Find the time when the particle is at the origin

The position function is given by x(t) = (t-2)^2. We need to find the time when x(t) = 0, which corresponds to the origin. Set x(t) equal to 0: \((t-2)^2 = 0\) Solve for t: \(t = 2\)

Differentiate the position function to obtain the velocity function

To find the velocity function, we need to differentiate the position function with respect to time, i.e., find the derivative of x(t) with respect to t. Use the chain rule to find the derivative: \(v(t) = \frac{dx(t)}{dt} = 2(t-2)(1) = 2(t-2)\)

Evaluate the velocity function at the time when the particle is at the origin

Now that we have the velocity function, we can evaluate it at the time when the particle is at the origin, which we found to be t = 2. v(2) = 2(2-2) = 2(0) = 0 m/s The velocity of the particle when it passes through the origin is 0 m/s, so the answer is (A).

Key concepts

Understanding the Position Function

The position function is a fundamental concept in particle motion. It describes the location of a particle at any given time. For this exercise, the position function given is \( x(t) = (t - 2)^2 \). This mathematical expression tells us how the position, denoted as \( x \), changes as time \( t \) progresses.

In simple terms, if you want to find out where the particle is at a particular moment, you replace \( t \) with a specific time value and solve the equation.

• For example, when \( t = 2 \), substituting into the function gives \( x(2) = (2 - 2)^2 = 0 \), meaning the particle is at the origin, where \( x = 0 \).

Remember, finding the time when the particle is at the origin is a crucial step in problems involving motion, and it usually involves setting the position function to zero and solving for \( t \). This step helps us understand when or if the particle reaches a certain position, which is essential in solving various motion problems.

Calculating the Velocity Function

Now, let's move on to the velocity function, which is derived from the position function. The velocity function

• indicates how fast a particle's position changes over time.
• In mathematical terms, this means differentiating the position function \( x(t) \) with respect to time \( t \).

The chain rule, a key differentiation technique, is often used during this process. It helps us differentiate composite functions like \( x(t) = (t - 2)^2 \). Applying the chain rule, you find the derivative as:\( v(t) = \frac{d}{dt}((t - 2)^2) = 2(t-2)(1) = 2(t-2) \)This new function, \( v(t) \), represents the particle's velocity at any given time \( t \). By plugging in different values of \( t \), you can determine how quickly the particle is moving and in what direction it's headed during its motion.

Using the Chain Rule for Differentiation

Differentiating complex functions is made simpler with the chain rule. This rule is especially useful for functions that appear as compositions of simpler, more straightforward functions. Understanding and applying the chain rule is an essential skill for solving many calculus problems.

Consider the position function \( x(t) = (t - 2)^2 \). Here, you see a function within a function

• the inner function is \( t - 2 \)
• the outer function is the squaring process, \(( ext{something})^2\).

To apply the chain rule:

• Step 1: Differentiate the outer function, treating the inner function as a single unit. This gives us \( 2(t-2) \).
• Step 2: Differentiate the inner function, which in this case is simple since the derivative of \( t-2 \) is \( 1 \).
• Step 3: Multiply the derivatives from Step 1 and Step 2 to compute the derivative of the full expression.

This approach offers a systematic way to tackle even complicated differentiation tasks, which is immensely helpful when working on physics problems involving motion like the one in this exercise. By distinguishing between functions within functions, calculus becomes more manageable and less daunting, allowing you to focus on the motion's intricate details.