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
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?
1 answer
1 answer