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) We could estimate that the instantaneous velocity at $t = 1$ is $6.28~m/s$
Work Step by Step
(a) We can find the height at $t=1$:
$y = 10(1)-1.86(1)^2 = 8.14~m$
(i) We can find the height at $t= 2$:
$y = 10(2)-1.86(2)^2 = 12.56~m$
We can find the average velocity:
$v_{ave} = \frac{12.56~m-8.14~m}{2~s-1~s} = 4.42~m/s$
(ii) We can find the height at $t= 1.5$:
$y = 10(1.5)-1.86(1.5)^2 = 10.815~m$
We can find the average velocity:
$v_{ave} = \frac{10.815~m-8.14~m}{1.5~s-1~s} = 5.35~m/s$
(iii) We can find the height at $t= 1.1$:
$y = 10(1.1)-1.86(1.1)^2 = 8.7494~m$
We can find the average velocity:
$v_{ave} = \frac{8.7494~m-8.14~m}{1.1~s-1~s} = 6.094~m/s$
(iv) We can find the height at $t= 1.01$:
$y = 10(1.01)-1.86(1.01)^2 = 8.202614~m$
We can find the average velocity:
$v_{ave} = \frac{8.202614~m-8.14~m}{1.01~s-1~s} = 6.2614~m/s$
(v) We can find the height at $t= 1.001$:
$y = 10(1.001)-1.86(1.001)^2 = 8.14627814~m$
We can find the average velocity:
$v_{ave} = \frac{8.14627814~m-8.14~m}{1.001~s-1~s} = 6.27814~m/s$
(b) As the time interval gets smaller, the average velocity gets closer to the instantaneous velocity at $t = 1$. We could estimate that the instantaneous velocity at $t = 1$ is $6.28~m/s$
