Ch 02: Motion Along a Straight Line

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 02: Motion Along a Straight Line

Problem 14

Chapter 2, Problem 14

A race car starts from rest and travels east along a straight and level track. For the first $5.0$ s of the car's motion, the eastward component of the car's velocity is given by $v_{x} (t) =$ ( $0.860$ m/s³)t². What is the acceleration of the car when $v sub x equals 12.0$ m/s?

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Identify the given velocity function: $ v_x(t) = 0.860 \, \text{m/s}^3 \cdot t^2 $. This function describes how the velocity of the car changes with time.

To find the acceleration, we need to differentiate the velocity function with respect to time. The acceleration $ a(t) $ is the derivative of $ v_x(t) $ with respect to $ t $.

Differentiate $ v_x(t) = 0.860 \, \text{m/s}^3 \cdot t^2 $ to find $ a(t) $. The derivative is $ a(t) = \frac{d}{dt}(0.860 \, \text{m/s}^3 \cdot t^2) = 2 \cdot 0.860 \, \text{m/s}^3 \cdot t = 1.72 \, \text{m/s}^3 \cdot t $.

We need to find the time $ t $ when the velocity $ v_x = 12.0 \, \text{m/s} $. Set $ 0.860 \, \text{m/s}^3 \cdot t^2 = 12.0 \, \text{m/s} $ and solve for $ t $.

Once $ t $ is found, substitute it back into the acceleration function $ a(t) = 1.72 \, \text{m/s}^3 \cdot t $ to find the acceleration at the moment when $ v_x = 12.0 \, \text{m/s} $.

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Key Concepts

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

Kinematics Equations

Kinematics equations describe the motion of objects without considering the forces that cause the motion. In this problem, the velocity function vx(t) = (0.860 m/s^3)t^2 is given, which allows us to find the acceleration by differentiating the velocity with respect to time.

Differentiation in Physics

Differentiation is a mathematical process used to find the rate at which a quantity changes. In physics, differentiating a velocity function with respect to time gives the acceleration. For vx(t) = (0.860 m/s^3)t^2, the acceleration a(t) is found by differentiating vx(t), resulting in a(t) = 2 * (0.860 m/s^3) * t.

Solving for Time

To find the acceleration at a specific velocity, we need to determine the time at which the velocity is 12.0 m/s. By setting vx(t) = 12.0 m/s and solving for t, we can substitute this time into the acceleration function a(t) to find the car's acceleration at that moment.

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Textbook Question

The table shows test data for the Bugatti Veyron Super Sport, the fastest street car made. The car is moving in a straight line (the $x$ -axis).

(a) Sketch a $v_{x}$ - $t$ graph of this car's velocity (in mi/h) as a function of time. Is its acceleration constant?

(b) Calculate the car's average acceleration (in m/s2) between (i) $0$ and $2.1$ s; (ii) $2.1$ s and $20.0$ s; (iii) $20.0$ s and $53$ s. Are these results consistent with your graph in part (a)? (Before you decide to buy this car, it might be helpful to know that only $300$ will be built, it runs out of gas in $12$ minutes at top speed, and it costs more than $dollar sign 1.5$ million!)

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Textbook Question

A physics professor leaves her house and walks along the sidewalk toward campus. After $5$ min, it starts to rain, and she returns home. Her distance from her house as a function of time is shown in Fig. E $2.10$ . At which of the labeled points is her velocity constant and negative?

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Textbook Question

A turtle crawls along a straight line, which we will call the $x$ -axis with the positive direction to the right. The equation for the turtle's position as a function of time is $x(t) = 50.0$ cm + ( $2.00$ cm/s) $t$ − ( $0.0625$ cm/s²) $t^2$ . At what time $t$ is the velocity of the turtle zero?

A turtle crawls along a straight line, which we will call the $x$ -axis with the positive direction to the right. The equation for the turtle's position as a function of time is $x(t) = 50.0$ cm + ( $2.00$ cm/s) $t$ − ( $0.0625$ cm/s²) $t^2$ . Sketch graphs of $x$ versus $t$ , $v_{x}$ versus $t$ , and $a_{x}$ versus $t$ , for the time interval $t equals 0$ to $t equals 40$ s.

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Textbook Question

A turtle crawls along a straight line, which we will call the $x$ -axis with the positive direction to the right. The equation for the turtle's position as a function of time is $x of t equals 50.0$ cm + ( $2.00$ cm/s) $t$ − ( $0.0625$ cm/s²) $t^{2}$ . Find the turtle's initial velocity, initial position, and initial acceleration.

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A race car starts from rest and travels east along a straight and level track. For the first 5.05.0 s of the car's motion, the eastward component of the car's velocity is given by vx(t)=v_{x}(t)= (0.8600.860 m/s3)t2. What is the acceleration of the car when vx=12.0v_{x}=12.0 m/s?

Verified video answer for a similar problem:

Key Concepts

Kinematics Equations

Differentiation in Physics

Solving for Time

A race car starts from rest and travels east along a straight and level track. For the first $5.0$ s of the car's motion, the eastward component of the car's velocity is given by $v_{x} (t) =$ ( $0.860$ m/s³)t². What is the acceleration of the car when $v sub x equals 12.0$ m/s?