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Ch 02: Motion Along a Straight Line
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 15th EditionCh 02: Motion Along a Straight LineProblema 14
Capitolo 2, Problema 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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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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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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