Problem 15 - Fundamentals of Physics

Question

(a) If a particlc's position is given by \(x=4-12 t+3 t^{2}\) (where \(t\) is in seconds and \(x\) is in meters), what is its velocity at \(t=1 \mathrm{~s} ?(\mathrm{~b})\) Is it moving in the positive or negative direction of \(x\) just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (c) Is there cver an instant when the velocity is zero? If so. give the time \(r\), if not, answer no. (f) Is there a time after \(t=3 \mathrm{~s}\) when the particle is moving in the ncgative dircction of \(x ?\) If so, give the time \(t\), if not, answer no.

Step-by-Step Solution

Answer

(a) -6 m/s (b) Negative (c) 6 m/s (d) Decreasing (e) Yes, t=2 s (f) No.

Step 1: Velocity Function

To find the velocity of the particle, we need to differentiate the position function with respect to time \(t\). Given the position function \(x = 4 - 12t + 3t^2\), differentiate it to get: \[v = \frac{dx}{dt} = -12 + 6t.\]

Step 2: Calculate Velocity at Specific Time

Substitute \(t = 1\) second into the velocity equation to find the velocity at this specific time:\[v = -12 + 6(1) = -6 \, \text{m/s}.\]

Step 3: Determine Direction of Motion

Since the velocity at \(t = 1\) second is \(-6 \, \text{m/s}\), the negative sign indicates the particle is moving in the negative direction of \(x\).

Step 4: Calculate Speed at Specific Time

Speed is the absolute value of velocity. Therefore, the speed at \(t = 1\) second is:\[|v| = |-6| = 6 \, \text{m/s}.\]

Step 5: Determine If Speed Is Increasing or Decreasing

The speed is increasing if the velocity and acceleration have the same sign. First, find the acceleration by differentiating the velocity function:\[a = \frac{dv}{dt} = 6 \, \text{m/s}^2.\]Since velocity (-6 m/s) is negative and acceleration (6 m/s^2) is positive, the speed is decreasing.

Step 6: Check for Zero Velocity

To find if there is a time when the velocity is zero, set the velocity equation to zero and solve for \(t\):\[-12 + 6t = 0 \t = 2 \, ext{seconds}.\]So, the velocity is zero at \(t = 2\) seconds.

Step 7: Check Negative Direction After Specific Time

Check if there is a time after \(t = 3\) seconds when the velocity is negative. Substitute a value greater than 3 into the velocity equation:\[v = -12 + 6(4) = 12 \, \text{m/s}.\]Thus, for any \(t > 3\) seconds, the velocity will be positive. Hence, there is no time after \(t = 3\) seconds when the particle moves in the negative direction of \(x\).

Key Concepts

Velocity

Velocity is a vector quantity that describes both the speed and direction of an object's motion. In this exercise, the velocity of a particle is derived from its position function. The position function given is \(x = 4 - 12t + 3t^2\).
The velocity, \(v\), is obtained by differentiating this position function with respect to time \(t\), which gives us \(v(t) = \frac{dx}{dt} = -12 + 6t\). By substituting \(t = 1\) into the velocity equation, we find the velocity at that specific time:

• \(v = -12 + 6(1) = -6\, \text{m/s}\) indicates the particle is moving with a velocity of \(-6\, \text{m/s}\).

This negative value reveals that at \(t = 1\) second, the particle is moving in the negative direction of the \(x\) axis.

Acceleration

Acceleration refers to the rate of change of velocity over time. It helps us understand how quickly an object is speeding up or slowing down. To find the acceleration from the velocity function, we differentiate again with respect to time \(t\). For the velocity function \(v(t) = -12 + 6t\), the acceleration \(a\) is calculated as:

• \(a = \frac{dv}{dt} = 6\, \text{m/s}^2\).

Here, the acceleration is a constant value of \(6\, \text{m/s}^2\).
This positive acceleration indicates that the particle is increasing in velocity over time. Even though at \(t = 1\) the velocity is negative, showing the particle is moving in the negative direction, the positive acceleration means the particle is slowing down its backward motion and will eventually start to move forward.

Position Function

The position function describes the location of a particle at any given time, \(t\). It is vital in kinematics for predicting where an object will be at any future time. For this problem, we're given the position function \(x = 4 - 12t + 3t^2\), where \(x\) is the position in meters, and \(t\) is the time in seconds.
This quadratic function suggests that the particle moves on a parabolic path over time. As \(t\) increases, the squared term \(3t^2\) becomes significant, influencing the shape of the path. By analyzing this function's derivatives (velocity and acceleration), we can build a full picture of the particle's motion over time. The points at which the velocity derivative equals zero are where the particle is momentarily at rest or changing direction.

Speed Analysis

Speed is the magnitude of velocity, essentially how fast an object is moving, regardless of direction. It is always expressed as a non-negative value. In this problem, speed at \(t = 1\) second is found by taking the absolute value of the velocity.

• Given \(v = -6\, \text{m/s}\), the speed is \(|v| = 6\, \text{m/s}\).

This means that regardless of direction, the particle is moving at a speed of \(6\, \text{m/s}\) at \(t = 1\) second.
To determine if the speed is increasing or decreasing, we evaluate the signs of velocity and acceleration. If they share the same sign, speed increases. If they have opposite signs, speed decreases. Here, the velocity is negative while acceleration is positive, resulting in decreasing speed at \(t=1\) second as the particle slows down in its backward (negative) motion.