Question

A turtle crawls along a straight line, which we will call the \(x\)-axis with the positive direction to the right. The equation for the turtle's position as a function of time is \(x(t) =\) 50.0 cm + (2.00 cm/s)\(t -\) (0.0625 cm/s\(^2)t^2\). (a) Find the turtle's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 cm from its starting point? What is the velocity (magnitude and direction) of the turtle at each of those times? (e) Sketch graphs of \(x\) versus \(t, v_x\) versus \(t\), and \(a_x\) versus \(t\), for the time interval \(t =\) 0 to \(t =\) 40 s.

Short answer

Initial conditions: Position = 50 cm, Velocity = 2 cm/s, Acceleration = -0.125 cm/s². Zero velocity at 16 s, returns to start at 32 s, distance 10 cm at 5.9, 21.1, 26.1, 37.1 s.

Step-by-step solution

Identifying Initial Conditions

The position function is given as \( x(t) = 50.0 \, \text{cm} + (2.00 \, \text{cm/s})t - (0.0625 \, \text{cm/s}^2)t^2 \). The initial position \( x(0) \) is simply the constant term, so \( x(0) = 50.0 \, \text{cm} \). The initial velocity is the coefficient of \( t \) in the position equation, \( v(0) = 2.00 \, \text{cm/s} \). Since the acceleration derives from the term involving \( t^2 \), the initial acceleration is \( a = -2 \times 0.0625 = -0.125 \, \text{cm/s}^2 \).

Finding Time When Velocity is Zero

The velocity as a function of time is found by differentiating the position equation: \( v(t) = \frac{d}{dt}[x(t)] = 2.00 - 2 \times 0.0625t = 2.00 - 0.125t \). Set the velocity equation to zero and solve for \( t \): \( 0 = 2.00 - 0.125t \Rightarrow t = \frac{2.00}{0.125} = 16 \, \text{s} \).

Time for Turtle to Return to Start

The turtle returns to its starting position when \( x(t) = 50.0 \, \text{cm} \). Setting the position function equal to 50: \( 50 + 2.00t - 0.0625t^2 = 50 \). Simplifying gives \( 0 = t(2.00 - 0.0625t) \). This equation yields two solutions: \( t = 0 \) and \( t = \frac{2.00}{0.0625} = 32 \, \text{s} \).

Time to Reach 10 cm from Starting Point

The distance from starting position is given by \( |x(t) - 50| = 10 \). Thus, solve \( |2.00t - 0.0625t^2| = 10 \). This translates to two cases: 1. \( 2.00t - 0.0625t^2 = 10 \): Solving \(0.0625t^2 - 2.00t + 10 = 0 \) gives roots via the quadratic formula (\( t \approx 5.9, 26.1 \)). 2. \( 2.00t - 0.0625t^2 = -10 \): Solving similarly gives \( t \approx 21.1, 37.1 \). Hence, two time intervals are valid: \( t = 5.9, 21.1, 26.1, 37.1 \) seconds.

Velocity at Specific Times

For each value of \( t \) found in the previous step (5.9s, 21.1s, 26.1s, 37.1s), calculate velocity using the velocity function \( v(t) = 2.00 - 0.125t \).For 5.9s: \( v = 2.00 - 0.125 \times 5.9 = 1.26 \, \text{cm/s} \).For 21.1s: \( v = 2.00 - 0.125 \times 21.1 = -0.64 \, \text{cm/s} \).For 26.1s: \( v = 2.00 - 0.125 \times 26.1 = -1.26 \, \text{cm/s} \).For 37.1s: \( v = 2.00 - 0.125 \times 37.1 = -2.64 \, \text{cm/s} \).

Sketching Graphs of Position, Velocity, and Acceleration

- **Position \( x(t) \)**: Plot \( x = 50 + 2.00t - 0.0625t^2 \) over [0,40] to see a parabola indicating initial increase followed by a decrease.- **Velocity \( v_x(t) \)**: Plot \( v = 2.00 - 0.125t \), a linear decreasing graph crossing 0 at 16s.- **Acceleration \( a_x(t) \)**: Constant line at \( -0.125 \, \text{cm/s}^2 \) since it's a constant acceleration.

Key concepts

Turtle Motion Across the Axis

When we think of turtle motion in kinematics, it is conceptualized through the idea of motion along an axis. In this exercise, a turtle moves along the x-axis in a straight line. This helps us understand movement in terms of linear motion, where we monitor changes in position over time.

In such problems, turtles, or any object for that matter, might start at a particular point, the initial position. For our turtle, it commences its journey at 50.0 cm along the x-axis. Motion can be characterized by position functions, which are mathematical functions describing how position changes as time elapses. This allows us to create a model and track where the turtle will be at any given moment.

We also delve into concepts like initial velocity and acceleration, which define how fast and in what manner (speeding up or slowing down) the turtle begins its journey. Initial velocity, here, is a forward speed of 2.00 cm/s, and acceleration is a deceleration of -0.125 cm/s², showing that after launching forward, the turtle gradually slows down.

Understanding Parabolic Trajectories

A parabolic trajectory in the context of this exercise refers to the type of path the turtle's motion takes due to its changing velocity. This path is not a physical 3D parabola, as with projectiles like thrown balls, but is evident in the position-time graph due to the acceleration acting on the turtle.

In the position function, the quadratic component \(0.0625 \text{ cm/s}^2 t^2\) gives this trajectory its parabolic nature. The trajectory here describes how the turtle moves along the x-axis as it initially speeds up and then slows down. The 'parabolic' aspect means the path rises to a peak and then diminishes, mimicking an upside-down arch when depicted on a graph.

For our turtle, this implies there's a specific moment where the speed ceases to increase and starts to decrease, leading to unique points such as where the turtle returns to its starting location. Understanding this helps predict motion patterns and changes over time, like when and where it pauses momentarily before shifting directions.

Role of Position Functions in Motion Analysis

In kinematics, a position function is vital for examining how the position of an object, like our turtle, changes over time. Mathematically, such a function provides a method to calculate the displacement from the starting point.For our turtle's motion, the position function is given as \(x(t) = 50.0 + 2.00t - 0.0625t^2\). Here's the breakdown:- **50.0 cm** represents the initial location of the turtle along the x-axis.- **2.00 cm/s** * t captures how the turtle moves in the positive x direction, akin to the turtle's initial speed pushing it forward.- **-0.0625 cm/s² \(t^2\)** showcases the negative acceleration, indicating the turtle is slowly coming to a stop as time progresses.Position functions allow for understanding not just where an object is but predict how an object will continue to move. By differentiating this function, we derive formulas for velocity, offering insights into speed changes over time, and further exploring temporal behavior of motion like pauses or returns to start.

Velocity Calculation Simplified

Velocity calculations are fundamental in understanding the real-time speed and direction of an object. In turtle motion analysis, it becomes crucial to compute to predict future positions accurately. From the given problem, velocity (e)) is merely the instantaneous rate of change of position.Obtained by differentiating the position function, we find the velocity function \(v(t) = 2.00 - 0.125t\). Consider these aspects:

• The constant term **2.00 cm/s** signifies that if the turtle maintained this constant rate, it would continue indefinitely in a linear path.
• The **-0.125t** represents how acceleration affects velocity, indicating a steady decrease in speed over time due to its deceleration.

Velocity is not static; it shifts as time passes which leads to dynamic motion scenarios like the point when the turtle's speed becomes zero providing insight into halts in motion and direction reversals. Such calculations are essential for determining moments of zero velocity, initial conditions, and adjustments in speed helping in understanding complete motion narratives.