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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 2 / Problem 53E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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CALC The acceleration of a motorcycle is given by a x(t) =

Chapter 2, Problem 53E

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QUESTION:

The acceleration of a motorcycle is given by \(a_{x}(t)=A t-B t^{2}\), where \(A=1.50 \mathrm{\ m} / \mathrm{s}^{3}\) and \(B=0.120 \mathrm{\ m} / \mathrm{s}^{4}\). The motorcycle is at rest at the origin at time t = 0. (a) Find its position and velocity as functions of time. (b) Calculate the maximum velocity it attains.

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QUESTION:

The acceleration of a motorcycle is given by \(a_{x}(t)=A t-B t^{2}\), where \(A=1.50 \mathrm{\ m} / \mathrm{s}^{3}\) and \(B=0.120 \mathrm{\ m} / \mathrm{s}^{4}\). The motorcycle is at rest at the origin at time t = 0. (a) Find its position and velocity as functions of time. (b) Calculate the maximum velocity it attains.

ANSWER:

Solution 53E

Part I

(a) The acceleration is given by

$\frac{{d}^{2}x(t)}{d{t}^{2}}=a(t)=At-B{t}^{2}$ … (1)

The velocity of the motorcycle can be obtained by integrating the above equation with respect to time $t$ .

Integrating equation (1) with respect to $t$ , we get

$v(t)=\frac{dx(t)}{dt}=\int\limits_{\ }^{\ }a(t)dt=\int\limits_{\ }^{\ }(At-B{t}^{2})dt=\frac{A{t}^{2}}{2}-\frac{B{t}^{3}}{3}+C$

Where $C$ is the constant of integration. Now, from the given problem we know that the motorcycle is at rest at time $t=0$ , hence putting this value in the above equation we have

$v(0)=0=C$

Hence the velocity as a function of $t$ can be written as

$v(t)=\frac{dx(t)}{dt}=\frac{1}{2}A{t}^{2}-\frac{1}{3}B{t}^{3}$ ….(2)

Now putting the value of $A$ and

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