Question

According to the drag equation the velocity of an object moving through a fluid can be modeled by the equation 2 dvkvdt=− where k is a constant. a) Find the general solution to this equation. b) An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. What will the velocity be after ten seconds?

Answer 1

The given differential equation is:

2dv/dt = -kv

where k is a constant.

a) To find the general solution to this differential equation, we can use separation of variables.

2dv/v = -k dt

Integrating both sides, we get:

2ln|v| = -kt + C1

where C1 is the constant of integration.

Taking exponential of both sides, we get:

|v|^2 = e^(C1) e^(-kt)

Since v can be either positive or negative, we can simplify the above equation as:

v = ±√(Ae^(-kt))

where A = e^(C1).

Therefore, the general solution to the given differential equation is:

v(t) = ±√(Ae^(-kt))

b) We are given that the initial velocity is 40 m/s and after 2 seconds, the velocity has decreased to 30 m/s. Let's use this information to find the value of A.

When t = 0, v = 40. Therefore,

40 = ±√(Ae^0)

40^2 = A

A = 1600

Now we can use the value of A to find the velocity after 10 seconds.

v(10) = ±√(1600e^(-10k))

We can use the information that the velocity has decreased to 30 m/s after 2 seconds to find the value of k.

30 = ±√(1600e^(-2k))

900 = 1600e^(-2k)

e^(-2k) = 0.5625

-2k = ln(0.5625)

k = -0.5 ln(0.5625)

Now we can substitute this value of k into the expression for v(10) to get:

v(10) = ±√(1600e^(5ln(0.5625)))

v(10) = ±√(1600(0.5625)^5)

v(10) = ±20.11 m/s

Therefore, the velocity of the object after ten seconds will be approximately 20.11 m/s.