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Unformatted text preview: e final velocity, not position, we can assume that the car was held in place at
y( t) = 0 until the force was applied and gravity was allowed to act on the car. That makes
x( t) = [200 − mg sin(θ )] u( t)
and the response is Solutions 3-45 M. J. Roberts - 8/16/04 − 1− e
y( t) = x( t) ∗ h( t) = [200 − mg sin(θ )] u( t) ∗
m t u( t) or
200 − mg sin(θ ) e m dτ
y( t) =
∫ 1 − kf
m − mf t m m − mf τ 200 − mg sin(θ ) 200 − mg sin(θ ) y( t) =
τ + e
t + k e
0 t The terminal velocity is the derivative of position as time approaches infinity which, in this
200 − mg sin(θ ) 200 − 2536.43
y′ ( +∞) =
Obviously a force of 200 N is insufficient to move the car forward and its terminal velocity is
negative indicating it is rolling backward down the hill.
62. A block of aluminum is heated to a temperature of 100 °C. It is then dropped into a
flowing stream of water which is held at a constant temperature of 10°C. After 10
seconds the temperature of the ball is 60°C. (Aluminum is such a good heat conductor
that its temperature is essentially uniform throughout its volume during the cooling
process.) The rate of cooling is proportional to the temperature difference between the
ball and the water.
Write a differential equation for this system with the temperature of the water as the
excitation and the temperature of the block as the response.
(b) Compute the time constant of the system. (c) Find the impulse response of the system and, from it, the step response. (d)
If the same block is cooled to 0 °C and dropped into a flowing stream of water at 80
°C, at time, t = 0, at what time will the temperature of the block reach 75°C?
(a) The controlling differential equation is
T ( t) = K ( Tw − Ta ( t))
dt a or 1d
T ( t) + Ta ( t) = Tw
K dt a
where Ta is the temperature of the aluminum ball and Tw is the temperature of the water.
(b) We can find the constant, K, by using the temperature after 10 seconds,
h( t) = Ke − Kt u( t) = 0.0588e −0.0588 t u( t) .
Solutions 3-46 M. J. Roberts - 8/16/04 (c) The unit step response is the integral of the impulse response,
h −1 ( t) = (1 − e −0.0588 t ) u( t) . 63. A well-stirred vat has been fed for a long time by two streams of liquid, fresh water at 0.2
cubic meters per second and concentrated blue dye at 0.1 cubic meters per second. The
vat contains 10 cubic meters of this mixture and the mixture is being drawn from the vat
at a rate of 0.3 cubic meters per second to maintain a constant volume. The blue dye is
suddenly changed to red dye at the same flow rate. At what time after the switch does the
mixture drawn from the vat contain a ratio of red to blue dye of 99:1?
Let the concentration of red dye be denoted by Cr ( t) and the concentration of blue
dye be denoted by Cb ( t) . The concentration of water is constant throughout at . The rates
of change of the dye concentrations are governed by
(VCb (t)) = −Cb (t) f draw
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This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.
- Spring '09