Diff Eq

Consider a body moving along a straight line, in the presence of an external force providing
constant acceleration a. In the absence of other forces, the velocity v of the body would increase
without bound if a 6= 0. But in reality, there will always be a decelerating force, due to friction,
air resistance, or viscous drag. This drag force increases in magnitude as the speed increases, and
as long as the speed is not too large, it is reasonable to assume direct proportionality. That is, in
addition to the constant acceleration a, the body experiences a deceleration proportional to velocity,
say, �v, where � is a positive constant of proportionality (the drag coe�cient). Since acceleration
equals rate of change of velocity, it follows that v_ = a 􀀀 �v, or equivalently,
v_ + �v = a:
(a) Given the body's initial velocity, v(0) = v0, derive a formula for the body's velocity, v = v(t),
as a function of time t. Qualitatively sketch the graph of v = v(t) in each of the relevant cases
(v0 < a
�, v0 = a
�, v0 > a
�) and give a physical interpretation of the results. What is the physical
signi�cance of the quantity a
�? What is the physical dimension of the drag coe�cient �, and what
is the physical signi�cance of its reciprocal, 1
�?
(b) Given the body's initial velocity, v(0) = v0, and initial position, x(0) = x0, derive a formula for
the body's position, x = x(t), as a function of time t. (Position is measured, relative to a reference
point x = 0, along the line of motion, in the same direction as v.) Hint: Use the formula for v
obtained in (a), along with the fact that velocity equals rate of change of position (that is, x_ = v)

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asked by Rina

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