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Physics & Further Mathematics: The velocity of an object is given by \(V=4t^3-3t^2+6t\). Find the value of \(t \) for which the acceleration is minimum.

 

The velocity of an object is given by V=4t^3-3t^2+6t. Find the value of t for which the acceleration is minimum.

Physics often presents intriguing questions about motion, velocity, and acceleration. Today, we’ll solve a problem where the velocity of an object is given by an equation.

Question

 The velocity of an object is given by \(V=4t^3-3t^2+6t\). Find the value of \(t \) for which the acceleration is minimum.

Solution

We are tasked with finding the value of t for which the acceleration is at a minimum. Let’s break it down step by step, explaining each differentiation rule applied for clarity.

Step 1: Understand the Relationship Between Velocity and Acceleration

Acceleration is the rate of change of velocity with respect to time. Mathematically, it is the first derivative of velocity (V) with respect to time (t):

\(Acceleration=\frac{Change \ in \ velocity}{time}\)

\(A(t)=\frac{\delta{V}}{\delta{t}}\)

 To determine when acceleration is at its minimum, we need to:

  1. Differentiate the velocity equation to find acceleration.
  2. differentiate the acceleration equation to find the rate of change of acceleration.
  3. Solve for \(t\)
  4. Verify that the result corresponds to a minimum acceleration.

Step 2: Differentiate the Velocity Equation to Find Acceleration

The given velocity equation is:

\(V=4t^3-3t^2+6t\)

Differentiation Rules Applied:

  • The derivative of \(t^n\) is \(n\times t^{n-1}\). This is called the power rule.
  • The derivative of a constant multiplied by a term is the constant multiplied by the derivative of the term.

Using these rules:

  1. Differentiate \(4t^3\): The power rule gives \(3\times 4t^{3-1}=12t^{2}\)
  2. Differentiate \(-3t^{2}\): The power rule gives \(2\times (-3)t^{2-1}=-6t\)
  3. Differentiate 6t: The power rule gives \(1\times 6t^{1-1}=6\)

Combining these results:

\(A(t)=12t^{2}-6t+6\)

This is the acceleration equation. Our goal from the question is to find when this acceleration reaches its minimum.

Step 3: Differentiate the Acceleration Equation

The minimum or maximum value of acceleration occurs when its rate of change (jerk) is zero. The jerk is the first derivative of acceleration with respect to time. Differentiate:

\(A(t)=12t^{2}-6t+6\)

Differentiation Rules Applied:

  •  Differentiate \(12t^{2}\): The power rule gives \(2\times 12t^{2-1}=24t\)
  • Differentiate \(-6t\): The power rule gives \(1\times (-6)t^{1-1}=-6\)
  • The derivative of a constant, 6 in this case, is 0.

Combining these results:

\(J(t)=24t-6\)

Step 4: Solve for \(t\) when \(J(t)=0\)

To find the critical point for acceleration, set the jerk to zero.

\(24t-6=0\)

Add 6 to both sides:

\(24t-6+6=0+6\)

\(24t=6\)

Divide both sides by 24

\(\frac{24t}{24}=\frac{6}{24}\)

 \(t=\frac{1}{4}\)

\(\therefore \) the value of \(t\) is \(\frac{1}{4}\)

Step 5: Verify That This Value of \(t\) Corresponds to a Minimum Acceleration

To confirm that \(t=\frac{1}{4}\) corresponds to a minimum acceleration, examine the sign of \(J(t)(24t-6)\) around \(t=\frac{1}{4}\):

  • For \(t<\frac{1}{4}\), \(J(t)\) is negative, meaning acceleration is decreasing.
  •  For \(t>\frac{1}{4}\), \(J(t)\) is positive, meaning acceleration is increasing.

Having gotten the value of  \(t\), let's go further and find the minimum acceleration. Although this was not asked in the question.

Step 6: Find the Minimum Acceleration

To find the minimum acceleration, substitute \(t=\frac{1}{4}\) into the acceleration equation:

\(A=12t^{2}-6t+6\)

Substitute \(t=\frac{1}{4}\):

\(A=12(\frac{1}{4})^{2}-6(\frac{1}{4})+6\)

\(A=12(\frac{1}{16})-\frac{6}{4}+6\)

\(A=\frac{12}{16}-\frac{3}{2}+6\)

\(A=\frac{3}{4}-\frac{3}{2}+6\)

\(A=\frac{3-6+24}{4}\)

\(A=\frac{21}{4}\)

\(A=5.25\)

\(\therefore\) the minimum acceleration is \(5.25 \ units\)

 

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