Problem 16 - Fundamentals of Physics

Question

The velocity \(\vec{v}\) of a particle moving in the \(x y\) plane is given by \(\vec{v}=\left(6.0 t-4.0 t^{2}\right) \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}}\), with \(\vec{v}\) in meters per second and \(t(>0)\) in seconds. (a) What is the acceleration when \(t=3.0 \mathrm{~s}\) ? (b) When (if ever) is the acceleration zero? (c) When (if ever) is the velocity zero? (d) When (if ever) does the speed equal \(10 \mathrm{~m} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

(a) \( -18.0 \mathrm{~m/s}^2 \hat{i} \); (b) \( t = 0.75 \) s; (c) \( t = 1.5 \) s; (d) \( t \approx 2.19 \) s.

Step 1: Find expression for acceleration

To find the acceleration, differentiate the velocity function with respect to time. The acceleration \( \vec{a} \) is the derivative of \( \vec{v}(t) \): \[ \vec{a} = \frac{d\vec{v}}{dt} = \left( \frac{d}{dt}(6.0 t - 4.0 t^2) \right) \hat{i} + \left( \frac{d}{dt}(8.0) \right) \hat{j} \]\[ = (6.0 - 8.0t) \hat{i} + 0 \hat{j} \] Thus, \( \vec{a}(t) = (6.0 - 8.0t) \hat{i}. \)

Step 2: Calculate acceleration at t=3.0 s

Substitute \( t = 3.0 \) seconds into the expression for acceleration:\[ \vec{a}(t) = (6.0 - 8.0 \times 3.0) \hat{i} = (6.0 - 24.0) \hat{i} = -18.0 \hat{i} \]So the acceleration at \( t = 3.0 \) seconds is \( -18.0 \mathrm{~m/s}^2 \hat{i} \).

Step 3: Determine when acceleration is zero

Set the acceleration function to zero and solve:\[ 6.0 - 8.0t = 0 \]\[ t = \frac{6.0}{8.0} = 0.75 \] Acceleration is zero at \( t = 0.75 \) seconds.

Step 4: Determine when velocity is zero

Set both components of the velocity to zero. For the \(x\)-component:\[ 6.0t - 4.0t^2 = 0 \]\[ t(6.0 - 4.0t) = 0 \]This gives \( t = 0 \) or \( t = 1.5 \). The \(y\)-component is constant at \(8.0 \), so it's never zero. Therefore, velocity is zero when \( t = 1.5 \) seconds.

Step 5: Determine when speed is 10 m/s

Speed is the magnitude of velocity.Calculate using:\[ \text{Speed} = \sqrt{(6.0t - 4.0t^2)^2 + 8.0^2} = 10 \]Squaring both sides:\[ (6.0t - 4.0t^2)^2 + 64 = 100 \]\[ (6.0t - 4.0t^2)^2 = 36 \]Taking square root:\[ 6.0t - 4.0t^2 = 6 \] or \[ 6.0t - 4.0t^2 = -6 \] Solving these gives two quadratic equations:1. \( 4.0t^2 - 6.0t + 6 = 0 \)2. \( 4.0t^2 - 6.0t - 6 = 0 \)The first has no real roots; solve the second:Using the quadratic formula,\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 4.0, b = -6.0, c = -6 \):\[ t = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot (-6)}}{8} \]\[ t = \frac{6 \pm \sqrt{36 + 96}}{8} \]\[ t = \frac{6 \pm \sqrt{132}}{8} \approx \frac{6 \pm 11.49}{8} \]\[ t \approx 2.19 \] (only considering positive time).Speed is \(10 \mathrm{~m/s} \) around \(t \approx 2.19 \) seconds.

Key Concepts

Velocity

Velocity is a fundamental concept in physics that describes the speed of a particle in a given direction. In the problem presented, the velocity of the particle moving in the xy-plane is given by a vector function: \( \vec{v} = (6.0t - 4.0t^2) \hat{i} + 8.0 \hat{j} \). Here, \(t\) represents time in seconds.

• The \( x \)-component, \( (6.0t - 4.0t^2) \hat{i} \), changes as time progresses, indicating the particle's motion in the horizontal direction.
• The \( y \)-component, \( 8.0 \hat{j} \), is constant, meaning the particle moves vertically at a steady speed.

It's crucial to remember that velocity is different from speed because it accounts for the direction of motion as well as the rate of change of position. Therefore, to find out when the particle's velocity reaches a particular state, you examine each component separately.

Acceleration

Acceleration is the rate of change of velocity with respect to time. It's a vector quantity; in this exercise, to find the acceleration, you need to differentiate the velocity function. The acceleration of the particle, \( \vec{a}(t) = (6.0 - 8.0t) \hat{i} \), shows how quickly the particle is speeding up or slowing down in the horizontal direction.

• When \( t = 3.0 \) seconds, substituting it into the acceleration expression gives \( \vec{a}(t) = -18.0 \mathrm{~m/s}^2 \hat{i} \), indicating deceleration.
• Acceleration is zero when \( 6.0 - 8.0t = 0 \), which occurs at \( t = 0.75 \) seconds, meaning the particle momentarily stops speeding up or slowing down.

Understanding when the acceleration is zero can help predict when a particle changes its motion type, for example, from speeding up to slowing down or vice versa.

Speed

Speed is a scalar quantity that represents how fast a particle is moving, regardless of direction. It's calculated using the magnitude of the velocity vector. In our case:
\[ \text{Speed} = \sqrt{(6.0t - 4.0t^2)^2 + 8.0^2} \]

• The speed is always non-negative and provides a single value for magnitude, unlike velocity which has both magnitude and direction.
• Solving when the speed is \( 10 \mathrm{~m/s} \) involves calculating \( \sqrt{(6.0t - 4.0t^2)^2 + 64} = 10\), leading to solving quadratic forms.

Understanding speed allows you to determine the total rate of the particle's travel, independent of its travel direction.

Quadratic Equations

Quadratic equations frequently appear in physics problems involving motion, such as determining when the speed reaches a specific value. The particle's motion leads to the quadratic form seen when determining when its speed is \( 10 \mathrm{~m/s} \):
\[ 4.0t^2 - 6.0t - 6 = 0 \]
Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we solve:

• The quadratic formula helps find the time values \( t \) when calculations are complex and involve squares.
• For the provided example, solving gives \( t \approx 2.19 \) seconds, indicating when the speed condition is met.

Quadratic equations are essential in problems involving changing velocities and accelerations, providing solutions where linear methods cannot suffice.