Answer to Question #88632 in Mechanics | Relativity for Megha

Question

The force of resistance on an object moving through a fluid is proportional to the square of its velocity. The initial velocity of an object of mass 1 kg moving through the fluid is 40 ms^−1. After 5 s the velocity reduces to 20 ms^−1. Calculate its velocity after
10 s.

Answer 1

$$ma=-kv^2$$

$$m\frac{dv}{dt}=-kv^2$$

$$\frac{dv}{v^2}=-\frac{k}{m} dt$$

$$\intop\frac{dv}{v^2}=-\frac{k}{m} \intop dt$$

$$-\frac{1}{v_0}=-\frac{k}{m} t +C$$

The initial condition:

$$-\frac{1}{40}=C\implies C=- 0.025$$

We have:

$$-\frac{1}{v}=-\frac{k}{m} t -0.025\implies \frac{1}{v}=\frac{k}{m} t+0.025$$

After 5 s:

$$\frac{1}{20}=\frac{k}{1} 5+0.025\implies k=0.005$$

After 10 s:

$$\frac{1}{v}=\frac{0.005}{1} 10+0.025$$

$$v=13.3\frac{m}{s}$$