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Which of the following 1D trajectories has a point t >0 where the velocity is O but the acceleration is not? O x(t) = -4t +1 O x(t) = In t O x(t) = 3e*/2 O x(t) = 4t + 9.8t2

Which of the following 1D trajectories has a point t >0 where the velocity is O but the acceleration is not? O x(t) = -4t +1 O x(t) = In t O x(t) = 3e*/2 O x(t) = 4t + 9.8t2

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Question:** Which of the following 1D trajectories has a point \( t \geq 0 \) where the velocity is 0 but the acceleration is not? **Options:** 1. \( x(t) = -4t + 1 \) 2. \( x(t) = \ln t \) 3. \( x(t) = 3e^{t^2/2} \) 4. \( x(t) = 4t + 9.8t^2 \) **Explanation:** We are given four different functions representing one-dimensional motion trajectories. We need to determine which of these trajectories has a point \( t \geq 0 \) where the velocity is zero, but the acceleration is not zero. To proceed: 1. Calculate the velocity for each trajectory by differentiating the trajectory function \( x(t) \) with respect to \( t \). 2. Set the velocity equal to zero and solve for \( t \). 3. For each solution, check the acceleration by differentiating the velocity function to see if it is non-zero at that point. **Solutions:** 1. For \( x(t) = -4t + 1 \): - Velocity \( v(t) = \frac{dx}{dt} = -4 \) - Acceleration \( a(t) = \frac{d^2x}{dt^2} = 0 \) - No point \( t \geq 0 \) where \( v(t) = 0 \) and \( a(t) \neq 0 \). 2. For \( x(t) = \ln t \): - Velocity \( v(t) = \frac{dx}{dt} = \frac{1}{t} \) - Acceleration \( a(t) = \frac{d^2x}{dt^2} = -\frac{1}{t^2} \) - \( v(t) = 0 \implies \frac{1}{t} = 0 \), which is impossible for \( t \geq 0 \). 3. For \( x(t) = 3e^{t^2/2} \): - Velocity \( v(t) = \frac{dx}{dt} = 3e^{t^2/2} \cdot t \) - Acceleration \( a(t) = \frac{d^
Transcribed Image Text:**Question:** Which of the following 1D trajectories has a point \( t \geq 0 \) where the velocity is 0 but the acceleration is not? **Options:** 1. \( x(t) = -4t + 1 \) 2. \( x(t) = \ln t \) 3. \( x(t) = 3e^{t^2/2} \) 4. \( x(t) = 4t + 9.8t^2 \) **Explanation:** We are given four different functions representing one-dimensional motion trajectories. We need to determine which of these trajectories has a point \( t \geq 0 \) where the velocity is zero, but the acceleration is not zero. To proceed: 1. Calculate the velocity for each trajectory by differentiating the trajectory function \( x(t) \) with respect to \( t \). 2. Set the velocity equal to zero and solve for \( t \). 3. For each solution, check the acceleration by differentiating the velocity function to see if it is non-zero at that point. **Solutions:** 1. For \( x(t) = -4t + 1 \): - Velocity \( v(t) = \frac{dx}{dt} = -4 \) - Acceleration \( a(t) = \frac{d^2x}{dt^2} = 0 \) - No point \( t \geq 0 \) where \( v(t) = 0 \) and \( a(t) \neq 0 \). 2. For \( x(t) = \ln t \): - Velocity \( v(t) = \frac{dx}{dt} = \frac{1}{t} \) - Acceleration \( a(t) = \frac{d^2x}{dt^2} = -\frac{1}{t^2} \) - \( v(t) = 0 \implies \frac{1}{t} = 0 \), which is impossible for \( t \geq 0 \). 3. For \( x(t) = 3e^{t^2/2} \): - Velocity \( v(t) = \frac{dx}{dt} = 3e^{t^2/2} \cdot t \) - Acceleration \( a(t) = \frac{d^
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