Answer
(a) An estimate for the distance traveled is $2.38~miles$ during this one minute time period.
(b) An estimate for the distance traveled is $2.36~miles$ during this one minute time period.
(c) We can not say that these estimates are lower estimates and upper estimates.
Work Step by Step
(a) The interval $[0,60]$ is divided into 6 subintervals.
$\Delta t = \frac{b-a}{n} = \frac{60~s-0}{6} = 10~s$
We can convert $\Delta t$ to units of hours:
$\Delta t = (10~s)\times \frac{1~h}{3600~s} = \frac{1}{360}~h$
To find an estimate for the distance, we can use the left endpoint of each subinterval:
$t_1 = 0$
$t_2 = 10$
$t_3 = 20$
$t_4 = 30$
$t_5 = 40$
$t_6 = 50$
We can find an estimate for the distance:
$\sum_{i=1}^{6} v(t_i)~\Delta t$
$= (182.9+168.0+106.6+99.8+124.5+176.1)~(\frac{1}{360})$
$= 2.38$
An estimate for the distance traveled is $2.38~miles$ during this one minute time period.
(b) To find another estimate for the distance, we can use the right endpoint of each subinterval:
$t_1 = 10$
$t_2 = 20$
$t_3 = 30$
$t_4 = 40$
$t_5 = 50$
$t_6 = 60$
We can find an estimate for the distance:
$\sum_{i=1}^{6} v(t_i)~\Delta t$
$= (168.0+106.6+99.8+124.5+176.1+175.6)~(\frac{1}{360})$
$= 2.36$
An estimate for the distance traveled is $2.36~miles$ during this one minute time period.
(c) Since the velocity function is not an increasing function or a decreasing function, we can not say that these estimates are lower estimates and upper estimates.
