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anonymous
 one year ago
The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2 sin πt + 4 cos πt, where t is measured in seconds. (Round your answers to two decimal places.)
(a) Find the average velocity during each time period.
(i) [1, 2]
cm/s
(ii) [1, 1.1]
cm/s
(iii) [1, 1.01]
cm/s
(iv) [1, 1.001]
cm/s
(b) Estimate the instantaneous velocity of the particle when t = 1.
cm/s
anonymous
 one year ago
The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2 sin πt + 4 cos πt, where t is measured in seconds. (Round your answers to two decimal places.) (a) Find the average velocity during each time period. (i) [1, 2] cm/s (ii) [1, 1.1] cm/s (iii) [1, 1.01] cm/s (iv) [1, 1.001] cm/s (b) Estimate the instantaneous velocity of the particle when t = 1. cm/s

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[s(t)=2\sin(\pi t)+4\cos(\pi t)\] We have the following formulae \[v_{avg}=\frac{\Delta s}{\Delta t}=\frac{s(t_{2})s(t_{1})}{t_{2}t_{1}}\] and \[v_{inst}=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}=\frac{ds}{dt}\] For your first interval (i) we have \[t_{1}=1s\]\[t_{2}=2s\] so you have to find \[v_{avg}=\frac{\Delta s}{\Delta t}=\frac{s(t_{2})s(t_{1})}{t_{2}t_{1}}=\frac{s(2)s(1)}{21}\] Finding s(2) and s(1) is as simple as plugging in values like you do for a function and the denominator can be solved simply as for the very last part of your question you'll have to find the derivative of the displacement with respect to time \[v_{inst}=\frac{ds}{dt}\] and evaluate at t=1 \[v_{inst}(1)=\frac{ds}{dt}_{t=1}\] Note that I've used v inst in "function" notation because it will also be a function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you even though my main problem is how to compute the values of t into the sin and cos equations.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[s(2)=2\sin(2\pi)+4\cos(2\pi)\] \[s(1)=2\sin(\pi)+4\cos(\pi)\] The value of sin(2pi) and cos(2pi) will be the same as sin(0) and cos(0) because if rotate through 2pi radians you'll come back to the same point dw:1443377341868:dw so for s2 we have \[s(2)=2\sin(0)+4\cos(0)\] and an interesting thing to look at is for example your (ii) part \[s(1.1)=2\sin(1.1\pi)+4\cos(1.1\pi)\] for something like this you should use a calculator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome thanks a lot!!
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