Calculus Problem

Question

A car traveling at 42 ft/sec decelerates at a constant 2 feet per second per second. How many feet does the car travel before coming to a complete stop?

______ feet

Answer 1

vo = 42 v(t) = integral of a(t) plus vo, the initial velocity

a(t) = -2

v(t) = -2t + 42 = 0 when t= 21 seconds, the time to stop

s(t) = -t^2 +42t

s(21) = -(21)^2 + 42(21) = 882-441 = 441 feet

1st second 41 feet (average velocity is (42+40)/2 = 41 the 1st second

2nd second 39 feet

3rd second 37

4th 31

...it's an arithmetic sequence with d=2, a1=42

nth 41-(n-1)d = 41-2(n-1)

21st 1 foot

sum 1 through 41 = 441