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Chapter 2: Problem 40
Find the distance traveled (to three decimal places) from \(t=1\) to \(t=5\)seconds, for a particle whose velocity is given by \(v(t)=t+\ln t .\) (A) 6.000 (B) 1.609 (C) 16.047 (D) 148.413
Short Answer
Expert verified
The answer is (C) 16.047.
Step by step solution
01
Identify the velocity function and the time interval
The velocity function is given by \(v(t)=t+\ln t .\) And we are given a time interval from \(t=1\) to \(t=5\) seconds.
02
Setup the Integral
Set up the definite integral of the absolute value of velocity from \(t=1\) to \(t=5\) seconds. Since the velocity function \(v(t)=t+\ln t\) is always increasing in the interval 1 to 5, we do not need to consider the absolute value. The integral then is: \(\int _1^5 v(t) dt\).
03
Evaluate the Integral
To evaluate this integral \(\int_1^5(t+\ln t) dt\), we can separate the two parts, reduce to the simpler integral of \(t\) and \(\ln(t)\) respectively, then use the fundamental theorem of calculus. The definite integral thus becomes: \(\int _1^5 t dt + \int _1^5 \ln(t) dt \). Each part can be calculated separately. The integral of \(t\) from 1 to 5 is \((1/2)*t^2]_1^5 = 12\) and the integral of \(\ln(t)\) from 1 to 5 is \(t*\ln(t) - t]_1^5 = (5 * \ln(5) -5) - (1 * \ln(1) -1) = 5 * \ln(5) -4\). Add them up, we get \(12 + 5 * \ln(5) -4 = 16.047.\)
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