Page 56 - How To Solve Word Problems In Calculus

3
2
(b) s = 2t − 27t + 84t + 25
ds 2
v = = 6t − 54t + 84
dt
When t = 1, s = 84 ft and v = 36 ft/min
When t = 5, s = 20 ft and v =−36 ft/min
When t = 10, s = 165 ft and v = 144 ft/min
dv
(c) a = = 12t − 54
dt
When t = 1, a =−42 ft/min 2
When t = 5, a = 6 ft/min 2
When t = 10, a = 66 ft/min 2
2
(d) From part (b) v = 6t − 54t + 84. Let v = 0 and solve for t.
2
0 = 6t − 54t + 84
0 = 6(t − 2)(t − 7)
t = 2 t = 7

The position of the object at these times is determined from the
position function. When t = 2, s = 101 and when t = 7,
s =−24.
(e) The object begins moving when t = 0. Its position at this time is
s = 25. During the first 2 minutes of travel (0 ≤ t ≤ 2), it moves
in the positive direction until its position s = 101. For the next 5
minutes (2 ≤ t ≤ 7) its velocity is negative so the object moves
in the negative direction until its position becomes s =−24. The
velocity then becomes positive again and for the next 3 minutes
(7 ≤ t ≤ 10) it moves in the positive direction until its position is
s = 165.
2
4. (a) h(t) = 80t − 16t + 96
h(0) = 96
h(2) = 192
h(2) − h(0) 192 − 96
v av = = = 48 ft/sec
2 − 0 2 − 0
(b) v(t) = h (t) = 80 − 32t

After 5 seconds the projectile
v(0) = 80 ft/sec has returned to the same height

(c) v(t) = h (t) = 80 − 32t from which it was thrown The
v(5) =−80 ft/sec velocity is negative since it is on
its way down.
h(5) = 96 ft

43