Problem 97 - Thomas Calculus

Question

Particle motion The position of a particle moving along a coordinate line is \(s=\sqrt{1+4 t},\) with \(s\) in meters and \(t\) in seconds. Find the particle's velocity and acceleration at \(t=6\) sec.

Step-by-Step Solution

Answer

The velocity at \( t=6 \) is \( \frac{2}{5} \) meters/second, and the acceleration is \( -\frac{2}{125} \) meters/second².

Step 1: Determine the Velocity Function

The velocity of a particle is the derivative of the position function with respect to time. The position function given is \( s(t) = \sqrt{1+4t} \). We use the chain rule to differentiate this function: \( v(t) = \frac{d}{dt} (1+4t)^{1/2} \). First, differentiate the outer function with respect to \( u = 1+4t \), which gives \( \frac{1}{2}(1+4t)^{-1/2} \). Then differentiate \( u \) with respect to \( t \), which gives \( 4 \). Hence, \( v(t) = \frac{1}{2}(1+4t)^{-1/2} \times 4 = \frac{2}{\sqrt{1+4t}} \).

Step 2: Calculate Velocity at t=6

Substitute \( t=6 \) into the velocity function \( v(t) = \frac{2}{\sqrt{1+4t}} \). This gives \( v(6) = \frac{2}{\sqrt{1+24}} = \frac{2}{\sqrt{25}} = \frac{2}{5} \) meters per second.

Step 3: Determine the Acceleration Function

The acceleration is the derivative of the velocity function with respect to time. We already have \( v(t) = \frac{2}{\sqrt{1+4t}} \). Differentiate this using the quotient rule. Let \( u(t) = 2 \) and \( v(t) = \sqrt{1+4t} \) with their derivatives as \( u'(t) = 0 \) and \( v'(t) = \frac{1}{2\sqrt{1+4t}} \times 4 \) respectively. Applying the quotient rule \( a(t) = \frac{(v(t)u'(t) - u(t)v'(t))}{(v(t))^2} = \frac{(\sqrt{1+4t}\times 0 - 2\times\frac{2}{2\sqrt{1+4t}})}{1+4t} = -\frac{2}{(1+4t)^{3/2}} \).

Step 4: Calculate Acceleration at t=6

Substitute \( t=6 \) into the acceleration function \( a(t) = -\frac{2}{(1+4t)^{3/2}} \). This gives \( a(6) = -\frac{2}{(1+24)^{3/2}} = -\frac{2}{125} \) meters per second squared.

Key Concepts

Particle Motion

Understanding particle motion is crucial in solving problems related to the movement of objects along a line over time. When we analyze the motion of a particle, we assess changes in its position, velocity, and acceleration. In the context of calculus, this involves differentiating position functions and interpreting the results in terms of real-world motion.
For a particle moving along a coordinate line, its position as a function of time, for example, \( s(t) = \sqrt{1+4t} \), helps us understand where it is at any given time. This kind of problem usually requires calculation of the velocity and acceleration of the particle
In summary:

• The position function describes where the particle is located in relation to time.
• Velocity and acceleration are derivatives of the position function, which provide insights into how fast and how the speed of the particle is changing.

Velocity Function

The velocity function represents how fast a particle is moving at any point in time. It is the first derivative of the particle's position function with respect to time. For instance, given the position function \(s(t) = \sqrt{1+4t}\), the velocity \(v(t)\) is found by taking its derivative.
To derive the velocity function, use the chain rule:

• Differentiating the outer function \( (1+4t)^{1/2} \) gives \( \frac{1}{2}(1+4t)^{-1/2} \).
• The inside function, \( 1+4t \), differentiates to 4.

Putting it together, we obtain \( v(t) = \frac{2}{\sqrt{1+4t}} \).
When we evaluate this at a specific time, say \( t=6 \), substituting gives \( v(6) = \frac{2}{5} \) meters per second, showing us that the particle's velocity is constant at this time.

Acceleration Function

Acceleration tells us how the velocity of a particle is changing over time. It's the derivative of the velocity function. In calculus, we find this by differentiating the velocity function again.
Given the velocity \( v(t) = \frac{2}{\sqrt{1+4t}} \), we use the quotient rule to find the acceleration function. The quotient rule is applied as:

• The numerator \( u(t) = 2 \) has a derivative of \( u'(t) = 0 \).
• The denominator \( v(t) = \sqrt{1+4t} \) has a derivative of \( v'(t) = \frac{2}{2\sqrt{1+4t}} \).

Applying the quotient rule produces the acceleration function \( a(t) = -\frac{2}{(1+4t)^{3/2}} \).
Substituting \( t=6 \) into this function, we get \( a(6) = -\frac{2}{125} \) meters per second squared, indicating the rate at which velocity decreases at that time.