Answer
(a) (i) 4.42 m/s
(ii) 5.35 m/s
(iii) 6.094 m/s
(iv) 6.2614 m/s
(v) 6.27814 m/s
(b) 6.28 m/s
Work Step by Step
The average velocity over the time interval $[t_{1},t_{2}]$ is equal to $\frac{y(t_{2})-y(t_{1})}{t_{2}-t_{1}}$. Therefore we can plug in respective values of $t_{1}$ and $t_{2}$ to find the average velocities.
(a) (i) $t_{1}$ = 1s, $t_{2}$ = 2s, $v_{avg}$ = 4.42 m/s
(ii) $t_{1}$ = 1s, $t_{2}$ = 1.5s, $v_{avg}$ = 5.35 m/s
(iii) $t_{1}$ = 1s, $t_{2}$ = 1.1s, $v_{avg}$ = 6.094 m/s
(iv) $t_{1}$ = 1s, $t_{2}$ = 1.01s, $v_{avg}$ = 6.2614 m/s
(v) $t_{1}$ = 1s, $t_{2}$ = 1.001s, $v_{avg}$ = 6.27814 m/s
(b) In part (a), it appears that the average velocities are approaching a value of 6.28 m/s as $t_{2}$ approaches 1s. So, an estimate of the instantaneous velocity when t=1 is 6.28 m/s.
