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Ch 02: Motion Along a Straight Line
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 2, Problem 14

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?

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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.
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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.
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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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Related Practice
Textbook Question

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. Draw vxv_x-tt and axa_x-tt graphs for the car's motion between t=0 t = 0 and t=5.00t = 5.00 s.

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Textbook Question

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a 1010-s interval. What are the magnitude, the algebraic sign, and the direction of the average acceleration in each interval? Assume that the positive direction is to the right.

(a) At the beginning of the interval, the astronaut is moving toward the right along the xx-axis at 15.015.0 m/s, and at the end of the interval she is moving toward the right at 5.05.0 m/s.

(b) At the beginning she is moving toward the left at 5.05.0 m/s, and at the end she is moving toward the left at 15.015.0 m/s.

(c) At the beginning she is moving toward the right at 15.015.0 m/s, and at the end she is moving toward the left at 15.015.0 m/s.

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Textbook Question

A turtle crawls along a straight line, which we will call the xx-axis with the positive direction to the right. The equation for the turtle's position as a function of time is x(t)=50.0x(t) = 50.0 cm + (2.002.00 cm/s)tt − (0.06250.0625 cm/s2)t2t^2. At what time tt is the velocity of the turtle zero?

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Textbook Question

A turtle crawls along a straight line, which we will call the xx-axis with the positive direction to the right. The equation for the turtle's position as a function of time is x(t)=50.0x(t) = 50.0 cm + (2.002.00 cm/s)tt − (0.06250.0625 cm/s2)t2t^2. Sketch graphs of xx versus tt, vxv_{x} versus tt, and axa_{x} versus tt, for the time interval t=0t = 0 to t=40t = 40 s.

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Textbook Question

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.

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Textbook Question

A turtle crawls along a straight line, which we will call the xx-axis with the positive direction to the right. The equation for the turtle's position as a function of time is x(t)=50.0x(t) = 50.0 cm + (2.002.00 cm/s)tt − (0.06250.0625 cm/s2)t2t^2. Find the turtle's initial velocity, initial position, and initial acceleration.

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