Question

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

Answer 1

\[s(t)=2\sin(\pi t)+4\cos(\pi t)\]
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

Answer 2

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

Answer 3

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

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