Answer:
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.