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)
To derive the formula for the body's velocity, v(t), as a function of time t, we can solve the differential equation v_ + �v = a.
We will use the method of integrating factors to solve this first-order linear ordinary differential equation. The integrating factor will be e^(∫�dt) = e^(�t), where t is the independent variable (time).
Multiplying both sides of the differential equation by the integrating factor, we get:
e^(�t) * (v_ + �v) = e^(�t) * a
Applying the product rule on the left side, we have:
(e^(�t) v)_ = e^(�t) * a
Integrating both sides with respect to t:
∫(e^(�t) v)_ dt = ∫(e^(�t) a) dt
Using the fundamental theorem of calculus, the left side becomes:
e^(�t) v = ∫(e^(�t) a) dt
Now, we can integrate the right side:
e^(�t) v = ∫(e^(�t) a) dt = (a/�) * e^(�t) + C
Where C is the constant of integration.
Dividing both sides by e^(�t), we get the final formula for the body's velocity:
v(t) = (a/�) + Ce^(-�t)
To determine the value of the constant C, we use the initial condition v(0) = v0. Substituting t = 0 and v = v0 into the formula:
v0 = (a/�) + Ce^(-�*0)
v0 = (a/�) + C
Solving for C:
C = v0 - (a/�)
Now, we have the complete formula for the body's velocity as a function of time:
v(t) = (a/�) + (v0 - (a/�)) * e^(-�t)
Now, let's analyze the formula and sketch the graph of v = v(t) in the relevant cases:
- For v0 < a/�, the term (v0 - (a/�)) will be negative. As time increases, the exponential term e^(-�t) will tend to zero, which means the velocity will approach the constant value of a/�.
- For v0 = a/�, the exponential term e^(-�t) will always be 1, and the velocity will remain constant at a/�.
- For v0 > a/�, the term (v0 - (a/�)) will be positive. As time increases, the exponential term e^(-�t) will tend to zero, but the velocity will still slowly decrease toward a/�.
The physical significance of the quantity a/� is the terminal velocity of the body. It represents the maximum velocity that the body can reach due to the opposing drag force. Once the body reaches terminal velocity, the net external force becomes zero, and the body will no longer accelerate.
The physical dimension of the drag coefficient � is [1/time], representing the rate at which the drag force increases with velocity. The reciprocal of the drag coefficient, 1/�, has the physical dimension of [time] and is often referred to as the characteristic time scale. It represents the time it takes for the velocity to decrease to 1/e (~37%) of its initial value due to the opposing drag force.
For the second part of the question, to derive a formula for the body's position, x = x(t), as a function of time t, we use the fact that velocity equals the rate of change of position (x_ = v). We already have the formula for v(t) obtained in part (a):
v(t) = (a/�) + (v0 - (a/�)) * e^(-�t)
To find x(t), we integrate both sides with respect to t:
∫v(t) dt = ∫((a/�) + (v0 - (a/�)) * e^(-�t)) dt
Integration gives:
x(t) = (a/�)t + (v0 - (a/�)) * (-1/�) * e^(-�t) + C'
Where C' is the constant of integration.
Using the initial condition x(0) = x0 (initial position), we can determine the value of the constant C':
x0 = (a/�) * 0 + (v0 - (a/�)) * (-1/�) * e^(-� * 0) + C'
x0 = -v0/� + C'
Solving for C':
C' = x0 + v0/�
The final formula for the body's position, x(t), as a function of time t, is:
x(t) = (a/�)t + (v0 - (a/�)) * (-1/�) * e^(-�t) + (x0 + v0/�)
Note: The position is measured relative to a reference point x = 0, along the line of motion, in the same direction as the velocity.