Math Solutions Sheet Page 2
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1
2
3(4.23) ≈ 53.72, and the
0.4 t
function x
(t) = 24.525 t + 61.3125 e
61.3125, so x
1
1
parachutist has covered a distance of 53.72 meters.
Second, after the parachute opens,
dv
2
= 9.81
1.2 v,
v
(0) = 20.
2
dt
1.2 t
The solution is v
(t) = 8.175 + 11.825 e
. We get the position function from
2
integrating v
and using x
(0) = 0:
2
2
1.2 t
x
(t) = 8.175 t
9.854 e
+ 9.854.
2
The parachutist reaches the ground when
t ≈ 236.9 sec.
x
(t) = 2000
53.72 = 1946.28
=⇒
2
Therefore, she reaches the ground after 4.23 + 236.9 ≈ 241 seconds.
13. When the velocity y of an object is very large, the magnitude of the force due
2
to air resistance is proportional to v
with the force acting in opposition to the
motion of the object. A shell of mass 3 kg is shot upward from the ground with
an initial velocity of 500 m/sec. If the magnitude of the force due to air resistance
2
is (0.1)v
, when will the shell reach its maximum height above the ground? What
is the maximum height?
Solution. The velocity of the shell is governed by the IVP
dv
1
2
=
9.81
v
,
v(0) = 500.
dt
30
The equation is not linear, so it must be solved as a separable equation. We get
v(t) = 17.16 tan(1.54
0.57 t).
The maximum height is reaches when
t ≈ 2.70 sec.
v(t) = 0
=⇒
To find the maximum height, first find the position function:
(
)
x(2.70) ≈ 104.78 m.
x(t) = 30.11 ln
cos(1.54
0.57 t)
+ 104.78
=⇒
2
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