Ch 02: Kinematics in One Dimension

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 02: Kinematics in One Dimension

Problem 39b

Chapter 2, Problem 39b

A particle's velocity is described by the function vₓ = (t² - 7t + 10) m/s, where t is in s. What is the particle's acceleration at each of the turning points?

Verified step by step guidance

To find the turning points of the particle's motion, we need to determine when the velocity is zero. Set the velocity function vₓ = t² - 7t + 10 equal to zero and solve for t. This gives the equation:

t ² - 7 t + 10 = 0

Solve the quadratic equation

t ² - 7 t + 10 = 0

using factoring, the quadratic formula, or another method. The solutions for t will give the times at which the velocity is zero, i.e., the turning points.

Next, calculate the acceleration of the particle at each turning point. Acceleration is the time derivative of velocity. Differentiate the velocity function vₓ = t² - 7t + 10 with respect to time t to find the acceleration function:

a = \frac{d v ₓ}{d t} = 2 t - 7

Substitute the values of t (from the turning points) into the acceleration function

a = 2 t - 7

to calculate the acceleration at each turning point.

Interpret the results. The calculated accelerations at the turning points indicate the rate of change of velocity at those specific times. This helps describe the particle's motion at the turning points.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Velocity

Velocity is a vector quantity that describes the rate of change of an object's position with respect to time. It includes both the speed of the object and the direction of its motion. In this context, the velocity function vₓ = (t² ─ 7t + 10) m/s indicates how the particle's speed changes over time, which is essential for determining its turning points.

Turning Points

Turning points occur when a particle changes direction, which corresponds to the points where its velocity is zero. To find these points, we set the velocity function equal to zero and solve for time t. Identifying these points is crucial for analyzing the particle's motion and calculating its acceleration at those specific moments.

Acceleration

Acceleration is the rate of change of velocity with respect to time and is a vector quantity. It can be derived from the velocity function by taking its derivative. In this case, calculating the acceleration at the turning points will provide insight into how the particle's speed is changing as it reverses direction, which is key to understanding its overall motion.

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Related Practice

Textbook Question

A particle's velocity is given by the function $v_{x} = (2.0 m/s) \sin (π t)$ , where $t$ is in $s$ . What is the first time after $t equals 0 s$ when the particle reaches a turning point?

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Textbook Question

A block is suspended from a spring, pulled down, and released. The block's position-versus-time graph is shown in FIGURE P2.38. At what times is the velocity zero? At what times is the velocity most positive? Most negative?

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Textbook Question

A particle's velocity is described by the function vₓ =kt² m/s, where k is a constant and t is in s. The particle's position at t₀ = 0 s is x₀ = -9.0 m. At t₁ = 3.0 s, the particle is at x₁ = 9.0 m. Determine the value of the constant k. Be sure to include the proper units.

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Textbook Question

A block is suspended from a spring, pulled down, and released. The block's position-versus-time graph is shown in FIGURE P2.38. Draw a reasonable velocity-versus-time graph.

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Textbook Question

The position of a particle is given by the function x = (2t³ - 6t² + 12) m, where t is in s. At what time is the acceleration zero?

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Textbook Question

A particle's velocity is given by the function $$\mathcal{v}$_x = (2.0 \, $\text{m/s}$) $\sin$($\pi$ t)$ , where $t$ is in $s$ . What is the particle's acceleration at that time?

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