A car's velocity as a function of time is given byvx(t)=α+βt2 v_x(t) = α + β$$t^2$$, where α=3.00α = 3.00 m/s and β=0.100β = 0.100 m/s3. Calculate the average acceleration for the time interval t=0t = 0 to t=5.00t = 5.00 s.

Ch 02: Motion Along a Straight Line

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 02: Motion Along a Straight Line

Problem 17a

Chapter 2, Problem 17a

A car's velocity as a function of time is given by $v_{x} (t) = α + β t^{2}$ , where $alpha equals 3.00$ m/s and $beta equals 0.100$ m/s³. Calculate the average acceleration for the time interval $t equals 0$ to $t equals 5.00$ s.

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Understand the problem: We need to calculate the average acceleration of a car over a given time interval using the velocity function v_x(t) = α + βt^2, where α = 3.00 m/s and β = 0.100 m/s^3.

Recall the formula for average acceleration: Average acceleration is defined as the change in velocity divided by the change in time. Mathematically, it is expressed as:

\frac{Δ v}{Δ t}

Calculate the initial and final velocities: Use the given velocity function v_x(t) = α + βt^2 to find the initial velocity at t = 0 and the final velocity at t = 5.00 s. Substitute t = 0 and t = 5.00 s into the equation to find v_x(0) and v_x(5).

Determine the change in velocity: Calculate the change in velocity by subtracting the initial velocity from the final velocity:

Δ v = v_x(5) - v_x(0)

Calculate the average acceleration: Use the formula for average acceleration with the calculated change in velocity and the time interval (5.00 s - 0 s) to find the average acceleration:

\frac{Δ v}{Δ t}

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Key Concepts

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

Velocity as a Function of Time

Velocity as a function of time describes how an object's speed and direction change over time. In this problem, the velocity is given by the equation v_x(t) = α + βt^2, where α is the initial velocity and βt^2 represents the change in velocity over time due to acceleration. Understanding this equation is crucial for determining how velocity evolves during the specified time interval.

Average Acceleration

Average acceleration is defined as the change in velocity divided by the time over which the change occurs. It provides a measure of how quickly an object's velocity changes on average over a given time period. In this problem, calculating the average acceleration involves finding the difference in velocity at the start and end of the time interval and dividing by the duration of the interval.

Integration of Velocity Function

To find the change in velocity over a time interval when given a velocity function, one can integrate the function with respect to time. In this context, integrating the velocity function v_x(t) = α + βt^2 from t = 0 to t = 5.00 s will yield the total change in velocity, which is necessary for calculating the average acceleration. This process involves applying basic calculus principles to solve the problem.

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Textbook Question

A car's velocity as a function of time is given by $v_x(t) = α + βt^2$ , where $α = 3.00$ m/s and $β = 0.100$ m/s³. Draw $v_{x}$ - $t$ and $a_{x}$ - $t$ graphs for the car's motion between $t equals 0$ and $t equals 5.00$ s.

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(a) At the beginning of the interval, the astronaut is moving toward the right along the $x$ -axis at $15.0$ m/s, and at the end of the interval she is moving toward the right at $5.0$ m/s.

(b) At the beginning she is moving toward the left at $5.0$ m/s, and at the end she is moving toward the left at $15.0$ m/s.

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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?

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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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The fastest measured pitched baseball left the pitcher's hand at a speed of $45.0$ m/s. If the pitcher was in contact with the ball over a distance of $1.50$ m and produced constant acceleration, what acceleration did he give the ball?

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A car's velocity as a function of time is given byvx(t)=α+βt2 v_x(t) = α + βt^2, where α=3.00α = 3.00 m/s and β=0.100β = 0.100 m/s3. Calculate the average acceleration for the time interval t=0t = 0 to t=5.00t = 5.00 s.

Verified video answer for a similar problem:

Key Concepts

Velocity as a Function of Time

Average Acceleration

Integration of Velocity Function

A car's velocity as a function of time is given by $v_{x} (t) = α + β t^{2}$ , where $alpha equals 3.00$ m/s and $beta equals 0.100$ m/s³. Calculate the average acceleration for the time interval $t equals 0$ to $t equals 5.00$ s.