Math 2280 - Assignment 3 Solutions

Math 2280 - Assignment 3 Solutions
Dylan Zwick
Fall 2013
Section 2.3 - 1, 2, 4, 10, 24
Section 2.4 - 1, 5, 9, 26, 30
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Section 2.3 - Acceleration-Velocity Models
2.3.1 The acceleration of a Maserati is proportional to the difference be-
tween 250 km/h and the velocity of this sports car. If the machine
can accelerate from rest to 100 km/h in 10s, how long will it take for
the car to accelerate from rest to 200 km/h?
Solution - The differential equation governing the car’s movement
will be:
dv
dt = k(250 − v).
This is a separable differential equation. We can rewrite it as:
dv
250 − v = kdt.
Integrating both sides of this equation we get:
∫ dv
250 − v =
∫
kdt.
⇒ − ln (250 − v) = kt + C
⇒ 250 − v = Ce−kt
⇒ v(t) = 250 − Ce−kt.
Using the initial condition v(0) = 0 = 250 − C we get C = 250. Using
the given v(10) = 100 we get:
2

v(10) = 250(1 − e−10k ) = 100
⇒ 1 − e−10k = 2
5
⇒ ln(e−10k) = ln
( 3
5
)
⇒ k = ln 5 − ln 3
10 ≈ .05108.
Using this value of k we want to find the value of t for which v(t) =
200. We do this by solving:
200 = 250(1 − e−.05108t∗ )
⇒ 4
5 = 1 − e−.05108t∗
⇒ e−.05108t∗ = 1
5
⇒ t∗ = ln 5
.05108 ≈ 31.5 seconds.
3

2.3.2 Suppose that a body moves through a resisting medium with resis-
tance proportional to its velocity v, so that dv/dt = −kv.
(a) Show that its velocity and position at time t are given by
v(t) = v0e−kt
and
x(t) = x0 +
( v0
k
)
(1 − e−kt).
(b) Conclude that the body travels only a finite distance, and find
that distance.
Solution
(a) - The differential equation
dv
dt = −kv
is separable, and can be rewritten as
dv
v = −kdt.
If we integrate both sides of the above differential equation we
get:
ln v = −kt + C
⇒ v(t) = Ce−kt.
Using the initial value v(0) = v0 = C we get:
v(t) = v0e−kt.
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