How do I incorporate the initial velocity into the equation?

In summary, the conversation discusses the problem of finding the time it takes for a car to accelerate from rest to 200 km/h, given that it can accelerate from rest to 100 km/h in 10 seconds. The conversation also includes an attempt to solve the problem using the equation dv/dt = k(250-v) and integrating it to get v = e^c * e^(-kt) - 250. However, there is confusion about how to equate the initial velocity (v0) into the equation.
  • #1
hydr
7
0
Ok here's my problem:

The acceleration of a car is proportional to the difference between 250 km/h and the velocity of the car. If this machine can accelerate from rest to 100 km/h in 10s, how long will it take for the car to accelerate from rest to 200 km/h?

Here is what I've done so far:

dv/dt = k(250-v)
integrating that...
[tex]\int[/tex]dv/(250-v) = [tex]\int[/tex]kdt which equates to
-ln|250-v| = kt + c solving for v gives me
v = e^c * e^(-kt) - 250

Now, i proceeded to solve for k, making e^c = 1 since i assumed c = vnaught and vnaught = 0

When i solved for t i ended up getting like 7.2 seconds which isn't right if it takes 10 seconds to accelerate to 100 km/h. My main question is: how do i equate vnaught into the equation? Because obviously i was mistaken to think vnaught = c.

I have also tried to make e^c = B, but once again am stuck as to how to equate B to vnaught. Any suggestions would be helpful, my book doesn't really explain much.
 
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  • #2
hydr said:
Now, i proceeded to solve for k, making e^c = 1 since i assumed c = vnaught and vnaught = 0
v0 is the initial velocity = 0. That's where t=0. That doesn't make e^c = 1 in any way.
 

1. What is a separable first order differential equation?

A separable first order differential equation is a type of differential equation that can be written in the form dy/dx = f(x)g(y), where f(x) and g(y) are functions of x and y respectively. This means that the variables x and y can be separated on opposite sides of the equation, making it easier to solve.

2. How do you solve a separable first order differential equation?

To solve a separable first order differential equation, you need to integrate both sides of the equation with respect to their respective variables. This will result in a general solution, which can then be solved for a specific solution by using initial conditions.

3. What are the applications of separable first order differential equations?

Separable first order differential equations are commonly used in physics, engineering, and other scientific fields to model various real-life phenomena. They can be used to model population growth, radioactive decay, and many other physical processes.

4. What is the difference between a separable first order differential equation and a non-separable one?

The main difference between a separable first order differential equation and a non-separable one is that the variables in a separable equation can be separated and solved independently, while in a non-separable equation, the variables are not easily separable and must be solved together.

5. Can a separable first order differential equation have multiple solutions?

Yes, a separable first order differential equation can have multiple solutions. This is because the general solution obtained by integrating both sides of the equation contains an arbitrary constant, which can result in different specific solutions when initial conditions are applied.

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