The problem involves a body falling under gravity with air resistance proportional to speed. Here's how we can model the situation and find the speed at the end of 5 seconds.

Given:

• Mass of the body, $m = 10 \, \text{kg}$
• Limiting speed, $v_{\text{lim}} = 50 \, \text{m/s}$
• Time, $t = 5 \, \text{seconds}$

The limiting speed occurs when the force of gravity is balanced by the air resistance, meaning: $mg = kv_{\text{lim}}$ where:

• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $k$ is the constant of proportionality for air resistance.

Step 1: Find $k$

From the equation $mg = kv_{\text{lim}}$, solve for $k$: $k = \frac{mg}{v_{\text{lim}}}$ Substitute the values: $k = \frac{10 \times 9.8}{50} = 1.96 \, \text{kg/s}$

Step 2: Differential equation for velocity

The equation of motion considering air resistance is: $m \frac{dv}{dt} = mg - kv$ Rearranging: $\frac{dv}{dt} = g - \frac{k}{m} v$ Substitute $\frac{k}{m} = \frac{1.96}{10} = 0.196$: $\frac{dv}{dt} = 9.8 - 0.196v$

Step 3: Solve the differential equation

This is a separable differential equation. Rewriting it: $\frac{dv}{9.8 - 0.196v} = dt$ Integrate both sides: $\int \frac{1}{9.8 - 0.196v} dv = \int dt$ The solution to this is: $-\frac{1}{0.196} \ln(9.8 - 0.196v) = t + C$ At $t = 0$, the body starts from rest ($v = 0$), so: $-\frac{1}{0.196} \ln(9.8) = C$ Thus, the solution becomes: $-\frac{1}{0.196} \ln(9.8 - 0.196v) = t - \frac{1}{0.196} \ln(9.8)$ Simplifying: $\ln \left( \frac{9.8 - 0.196v}{9.8} \right) = -0.196t$ Exponentiating both sides: $\frac{9.8 - 0.196v}{9.8} = e^{-0.196t}$ Solving for $v$: $v = 50 \left( 1 - e^{-0.196t} \right)$

Step 4: Find the speed at $t = 5$ seconds

Substitute $t = 5$ into the equation for $v$: $v = 50 \left( 1 - e^{-0.196 \times 5} \right)$ Let's calculate this.The speed of the body at the end of 5 seconds is approximately $31.23 \, \text{m/s}$.

Would you like more details on any part of the process?

Here are 5 related questions to expand on this concept:

1. How would the speed change if the limiting speed was different?
2. What would happen to the speed if air resistance was not proportional to speed but to speed squared?
3. How can the same problem be modeled for an object falling in a vacuum?
4. How does the mass of the body affect the limiting speed and final velocity?
5. What is the terminal velocity, and how is it connected to the limiting speed?

Tip: When solving problems with air resistance, identifying the terminal or limiting speed helps simplify the equations significantly.