Answer
(a) $V'(5)$ = $-218.75$ $gallons/min$
(b) $V'(10)$ = $-187.5$ $gallons/min$
(c) $V'(20)$ = $-125$ $gallons/min$
(d) $V'(40)$ = $0$ $gallons/min$
the rate is fastest when t = 0 (at t = 0 the rate is -250 gallons/minute) while it is the slowest at t = 40 (at t = 40, the rate is zero, in other words the water is all gone)
Work Step by Step
$V(t)$ = $5000(1-\frac{1}{40}t)^{2}$
$V'(t)$ = $-250(1-\frac{1}{40}t)$
(a) $V'(5)$ = $-250[1-\frac{1}{40}(5)]$ = $-218.75$ $gallons/min$
(b) $V'(10)$ = $-250[1-\frac{1}{40}(10)]$ = $-187.5$ $gallons/min$
(c) $V'(20)$ = $-250[1-\frac{1}{40}(20)]$ = $-125$ $gallons/min$
(d) $V'(40)$ = $-250[1-\frac{1}{40}(40)]$ = $0$ $gallons/min$
the rate is fastest when t = 0 (at t = 0 the rate is -250 gallons/minute) while it is the slowest at t = 40 (at t = 40, the rate is zero, in other words the water is all gone)
