## 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

1. anonymous

$s(t)=2\sin(\pi t)+4\cos(\pi t)$ 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 $v_{avg}=\frac{\Delta s}{\Delta t}=\frac{s(t_{2})-s(t_{1})}{t_{2}-t_{1}}=\frac{s(2)-s(1)}{2-1}$ 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

2. anonymous

Thank you even though my main problem is how to compute the values of t into the sin and cos equations.

3. anonymous

$s(2)=2\sin(2\pi)+4\cos(2\pi)$ $s(1)=2\sin(\pi)+4\cos(\pi)$ 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

4. anonymous

Awesome thanks a lot!!