Question

The position of a particle moving along the \(x\) -axis varies with time according to the expression \(x=4 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. Evaluate the particle's position a) at \(t=2.00 \mathrm{~s}\). b) at \(2.00 \mathrm{~s}+\Delta t\) c) Evaluate the limit of \(\Delta x / \Delta t\) as \(\Delta t\) approaches zero, to find the velocity at \(t=2.00 \mathrm{~s}\).

Short answer

Answer: The particle's position at t = 2.00 s is 16 meters, and its velocity at that time is 16 m/s.

Step-by-step solution

Part (a): Calculate the position at t = 2.00 s

To find the position of the particle at \(t = 2.00\) s, plug the value of \(t\) into the given expression for position \(x\): \(x = 4t^2\) \(x(2.00) = 4(2.00)^2\) \(x(2.00) = 4(4)\) \(x(2.00) = 16 \, \text{m}\) The particle's position at \(t = 2.00\) s is 16 meters.

Part (b): Calculate the position at t = 2.00 s + Δt

To find the position at \(t = 2.00 \, \text{s} + \Delta t\), plug this value of \(t\) into the position function: \(x(t) = 4t^2\) \(x(2.00+\Delta t) = 4(2.00+\Delta t)^2\) The position of the particle at \(t = 2.00\,\text{s} + \Delta t\) is \(x(2.00+\Delta t) = 4(2.00+\Delta t)^2\).

Part (c): Evaluate the limit of Δx/Δt to find the velocity at t = 2.00 s

To find the velocity of the particle at \(t = 2.00\) s, we need to find the limit of the difference quotient as \(\Delta t\) approaches zero: \(\lim_{\Delta t \to 0} \frac{x(2.00+\Delta t) - x(2.00)}{\Delta t}\) First, let's simplify the expression numerator: \(x(2.00+\Delta t) - x(2.00) = 4(2.00+\Delta t)^2 - 4(2.00)^2\) \(= 4[(2.00+\Delta t)^2 - (2.00)^2]\) Next, factor the difference of squares in the brackets: \(= 4[(2.00+\Delta t - 2.00)(2.00+\Delta t + 2.00)]\) \(= 4[\Delta t(4.00+\Delta t)]\) Now, let's put this back into our limit: \(\lim_{\Delta t \to 0} \frac{4[\Delta t(4.00+\Delta t)]}{\Delta t}\) We can cancel the \(\Delta t\) terms: \(\lim_{\Delta t \to 0} 4(4.00+\Delta t)\) Now, take the limit as \(\Delta t\) approaches zero: \(4(4.00 + 0)\) \(= 4(4.00)\) \(= 16 \, \mathrm{m/s}\) The particle's velocity at \(t = 2.00\) s is 16 m/s.

Key concepts

Position Function

In kinematics, the position function describes how a particle's location changes over time. For a particle moving along a straight line, this function is mathematically represented. In our exercise, the position function is given by the equation:\[ x(t) = 4t^2 \]Here, \(x\) is the particle's position along the x-axis in meters, and \(t\) represents time in seconds. This specific function tells us that the position of the particle depends on the square of the time elapsed. As time increases, the particle's position changes quadratically, not linearly.
To find the particle's position at any given time, we simply substitute the time value into this expression. For example, at \(t = 2\) seconds, substituting into the position function gives:\[x(2) = 4(2)^2 = 16 \text{ meters} \]

Velocity

Velocity is a crucial concept in understanding motion. It tells us how fast and in what direction a particle's position is changing. Unlike speed, which only considers the magnitude, velocity is a vector quantity. In the context of our position function, velocity can be understood as the slope of the position-time graph at any particular moment.
Mathematically, we often determine velocity using the derivative of the position function, as this finds the rate at which position changes with respect to time. In our exercise, the velocity at \(t = 2.00\) seconds can be found by evaluating the limit as the change in time approaches zero. This gives us the instantaneous velocity:- At \(t = 2.00\) seconds, the calculated velocity is 16 m/s.

Difference Quotient

The difference quotient is a foundational concept used to find the average rate of change of a function over an interval. It is represented as the ratio of the change in the function's value to the change in the input. In simpler terms, it's a way to compute how much the output of a function changes per unit change in the input.
For our position function, the difference quotient helps in finding the average velocity over a time interval \(\Delta t\). It is expressed as\[ \frac{x(t + \Delta t) - x(t)}{\Delta t} \]Using this with our position function \(x(t) = 4t^2\), the expression simplifies, helping us to find approximate velocity over small time changes. However, for instantaneous velocity, we need to take the limit as \(\Delta t\) approaches zero.

Limit

Limits are a fundamental concept when understanding changes at very small scales, particularly in calculus. They allow us to determine the behavior of functions as they approach specific points or boundaries. In kinematics, limits are essential for finding instantaneous rates of change, like velocity.In our exercise, to find the velocity of the particle at exactly \(t = 2.00\) seconds, we calculated the limit of the difference quotient as \(\Delta t\) approaches zero:\[ \lim_{\Delta t \to 0} \frac{x(2.00 + \Delta t) - x(2.00)}{\Delta t} \]This demonstrated the velocity at that instant, resulting in a value of 16 m/s. By using limits, we transition from average rates over intervals to precise, instantaneous values, capturing the exact speed and direction at a given moment.