Newton's equation for a free falling object with air resistance of mass m kilograms says that its velocity v(t) satisfies the DE
mv'(t) = mg - kv(t)
where g = 9.8 \frac{m}{s^2} and v(t) is measured in \frac{m}{s}. Suppose that k=2 \frac{kg}{sec} and that a ball of mass m = 1 kg falls with initial velocity v(0)=0. Round to the nearest tenth.
(a) What is its velocity after 1.5 seconds?
So far I have this, but not sure where to go from this:
dv/dt + 2v = 9.8
(b) What is the terminal velocity?
Your DE yields
v = c e^(-2t) + 4.9
v(0) = 0 means c = -4.9, so
v(t) = 4.9 - 4.9e^(-2t)
(a) plug in t=1.5
(b) v(∞) = 4.9
To find the velocity of the object after 1.5 seconds, we need to solve the differential equation:
mv'(t) = mg - kv(t)
First, let's rewrite the equation in terms of acceleration (a) instead of velocity (v):
ma = mg - kv(t)
Since a = v'(t), we can rearrange the equation as:
mv'(t) + kv(t) = mg
Now, substitute the given values of m = 1 kg, k = 2 kg/sec, and g = 9.8 m/s² into the equation:
v'(t) + 2v(t) = 9.8
This is a first-order linear ordinary differential equation (ODE) in the form of:
v'(t) + p(t)v(t) = q(t)
where p(t) = 2 and q(t) = 9.8.
To solve this type of equation, we can use an integrating factor. The integrating factor (IF) is given by:
IF = e^(∫p(t)dt) = e^(∫2dt) = e^(2t)
Now, multiply both sides of the equation by the integrating factor:
e^(2t)v'(t) + 2e^(2t)v(t) = 9.8e^(2t)
Apply the product rule on the left side to simplify:
(d/dt)(e^(2t)v(t)) = 9.8e^(2t)
Integrate both sides with respect to t:
∫(d/dt)(e^(2t)v(t)) dt = ∫9.8e^(2t) dt
e^(2t)v(t) = ∫9.8e^(2t) dt
Integrating the right side of the equation gives:
e^(2t)v(t) = 4.9e^(2t) + C
To find the constant of integration (C), we need to use the initial condition v(0) = 0:
e^(2(0))v(0) = 4.9e^(2(0)) + C
v(0) = 0 = 4.9 + C
C = -4.9
Substitute the value of C back into the equation:
e^(2t)v(t) = 4.9e^(2t) - 4.9
Finally, isolate v(t):
v(t) = (4.9e^(2t) - 4.9) / e^(2t)
Now, we can substitute t = 1.5 seconds into this equation to find the velocity after 1.5 seconds:
v(1.5) = (4.9e^(2(1.5)) - 4.9) / e^(2(1.5))
Evaluate this expression using a calculator or software to get the velocity after 1.5 seconds.
For part (b), the terminal velocity occurs when the velocity (v) becomes constant, meaning v'(t) = 0. Rearranging the original differential equation, we have:
ma = mg - kv(t)
0 = mg - kv(t)
Solving for v(t), we get:
v(t) = mg / k
Substituting the given values of m = 1 kg, g = 9.8 m/s², and k = 2 kg/sec into the equation:
v(t) = (1 kg)(9.8 m/s²) / (2 kg/sec)
Evaluate this expression to find the terminal velocity.