Ch 03: Motion in Two or Three Dimensions

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 03: Motion in Two or Three Dimensions

Problem 8b

Chapter 3, Problem 8b

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by $\vec{v}\) = $\left$[ 5.00~$\mathrm{m/s}$ - (0.0180~$\mathrm{m/s^3}$)t^2 $\right$] $\hat{i}$ + $\left$[ 2.00~$\mathrm{m/s}$ + (0.550~$\mathrm{m/s^2}$)t $\right$] \(\hat{j}$ . What are the magnitude and direction of the car's velocity at $t equals 8.00 s$ ? (b) What are the magnitude and direction of the car's acceleration at $t equals 8.00 s$ ?

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To find the velocity of the car at t = 8.00 s, substitute t = 8.00 s into the given velocity function v(t) = [5.00 m/s − (0.0180 m/s³)t²]î + [2.00 m/s + (0.550 m/s²)t]ĵ. Calculate the components separately: v_x = 5.00 m/s − (0.0180 m/s³)(8.00 s)² and v_y = 2.00 m/s + (0.550 m/s²)(8.00 s).

Calculate the magnitude of the velocity vector using the Pythagorean theorem: |v| = √(v_x² + v_y²). Substitute the values of v_x and v_y obtained from the previous step.

Determine the direction of the velocity vector by calculating the angle θ with respect to the positive x-axis using the tangent function: θ = arctan(v_y / v_x).

To find the acceleration of the car at t = 8.00 s, differentiate the velocity function with respect to time to get the acceleration function: a(t) = d/dt[v(t)]. The x-component of acceleration, a_x, is the derivative of v_x(t) = 5.00 m/s − (0.0180 m/s³)t², and the y-component, a_y, is the derivative of v_y(t) = 2.00 m/s + (0.550 m/s²)t.

Substitute t = 8.00 s into the acceleration components a_x and a_y to find the acceleration at that time. Calculate the magnitude of the acceleration vector using |a| = √(a_x² + a_y²) and determine the direction using θ = arctan(a_y / a_x).

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Key Concepts

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

Velocity Vector

Velocity is a vector quantity that describes the rate of change of position with respect to time. It has both magnitude and direction. In this problem, the velocity is given as a function of time with components in the î and ĵ directions, representing motion in a two-dimensional plane. Understanding how to calculate the magnitude and direction from these components is crucial.

Acceleration Vector

Acceleration is the rate of change of velocity with respect to time, also a vector quantity. It can be derived by differentiating the velocity function with respect to time. The problem requires calculating the acceleration at a specific time, which involves finding the derivative of each component of the velocity vector and evaluating it at t = 8.00 s.

Vector Magnitude and Direction

The magnitude of a vector is calculated using the Pythagorean theorem, combining its components. The direction is typically expressed as an angle relative to a reference axis, found using trigonometric functions like tangent. For this problem, determining the magnitude and direction of both velocity and acceleration vectors at a specific time involves these calculations.

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A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by $\vec{v}\) = $\left$[ 5.00~$\mathrm{m/s}$ - (0.0180~$\mathrm{m/s^3}$)t^2 $\right$] $\hat{i}$ + $\left$[ 2.00~$\mathrm{m/s}$ + (0.550~$\mathrm{m/s^2}$)t $\right$] \(\hat{j}$ . What are $a_{x} (t)$ and $a_{y} (t)$ , the $x$ - and $y$ - components of the car's velocity as functions of time?

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