Answer
(a) (i) 6 cm/s
(ii) -4.712 cm/s
(iii) -6.134 cm/s
(iv) -6.268 cm/s
(b) $-2\pi$ cm/s
Work Step by Step
The average velocity of the particle from time $t_{1}$ to $t_{2}$ is equal to $\frac{s(t_{2})-s(t_{1})}{t_{2}-t_{1}}$. We can then plug in values of $t_{1}$ and $t_{2}$ for each of the questions below.
(a) (i) $t_{1}$ = 1s, $t_{2}$ = 2s, $v_{avg}$ = 6 cm/s
(ii) $t_{1}$ = 1s, $t_{2}$ = 1.1s, $v_{avg}$ = -4.712 cm/s
(iii) $t_{1}$ = 1s, $t_{2}$ = 1.01s, $v_{avg}$ = -6.134 cm/s
(iv) $t_{1}$ = 1s, $t_{2}$ = 1.001s, $v_{avg}$ = -6.268 cm/s
(b) We can essentially take the first derivative of the equation in the question at t = 1.
$s = 2 sin(\pi t) + 3 cos(\pi t)$
$s' = 2\pi cos(\pi t) - 3\pi sin(\pi t)$
$s'(1) = 2\pi cos(\pi) - 3\pi sin(\pi)$
$s'(1) = -2\pi$
Thus the instantaneous velocity at t=1 is $-2\pi$ cm/s.
We can also observe that, in part (a), as $t_{2}$ approaches 1s, the velocity approaches $-2\pi$ cm/s.
