Problem 24

[Pages:3]Boyce & DiPrima ODEs 10e: Section 2.3 - Problem 24

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Problem 24

A rocket sled having an initial speed of 150 mi/h is slowed by a channel of water. Assume that during the braking process, the acceleration a is given by a(v) = -?v2, where v is the velocity and ? is a constant.

(a) As in Example 4 in the text, use the relation dv/dt = v(dv/dx) to write the equation of motion in terms of v and x.

(b) If it requires a distance of 2000 ft to slow the sled to 15 mi/h, determine the value of ?.

(c) Find the time required to slow the sled to 15 mi/h.

Solution Part (a)

The acceleration is given by

a = -?v2.

Replace a with dv/dt.

dv = -?v2 dt

Use the chain rule to write v in terms of x:

dv dv dx dv

=

= v.

dt dx dt dx

As a result, the previous equation becomes

dv v = -?v2. dx

Divide both sides by v2

dv

dx = -? v

and then rewrite the left side as a derivative of a logarithm.

d ln |v| = -?

dx

If we take the positive x-axis to point in the direction the sled is moving, then the absolute value sign can be dropped because v will be positive. Integrate both sides with respect to x.

ln v = -?x + C

Exponentiate both sides.

v(x) = e-?x+C = eC e-?x

Use a new constant A for eC.

v(x) = Ae-?x



Boyce & DiPrima ODEs 10e: Section 2.3 - Problem 24

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Use the fact that the initial speed is 150 mi/h to determine A.

v(0) = A = 150

So then Part (b)

v(x) = 150e-?x.

Now use the fact that the sled requires a distance of 2000 ft to slow to 15 mi/h. Set v(x) = 15 and

x = 2000 and solve for ?.

15 = 150e-2000?

e-2000? = 1 10

ln e-2000? = ln 1 10

-2000? = - ln 10

Therefore,

ln 10 1 5280 ft 66

1

1

?=

?

= ln 10 6.08

2000 ft mi 25

mi

mi

66 ln 10

v(x) = 150 exp -

x,

25

where x is in miles and v is in miles per hour.

Part (c)

Replace v(x) with dx/dt and solve the resulting ODE by separating variables.

dx

66 ln 10

= 150 exp -

x

dt

25

66 ln 10

exp

x dx = 150 dt

25

Integrate both sides.

25

66 ln 10

exp 66 ln 10

25 x = 150t + C2

Apply the initial condition x(0) = 0 to determine C2.

25 66 ln 10 = C2

The previous equation is then

25

66 ln 10

25

exp

x = 150t +

.

66 ln 10

25

66 ln 10

Solve for t.

25

66 ln 10

150t =

exp

x -1

66 ln 10

25



Boyce & DiPrima ODEs 10e: Section 2.3 - Problem 24

Divide both sides by 150.

1

66 ln 10

t=

exp

x -1 .

396 ln 10

25

Therefore, the time it takes for the sled to go

1 mi 25

x = 2000 ft ?

= mi

5280 ft 66

is

1

3600 seconds

=

h?

35.53 seconds.

44 ln 10

1 hour

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