Solve a Particle's Velocity Given Acceleration: Calculus Task

In summary, the conversation discusses solving a separable differential equation to find the velocity of a particle with an initial velocity of 5m/s and an acceleration of a=-.5v. Through integration and using the initial condition of v(0)=5, the solution is found to be v=5e^(-1/2 t).
  • #1
vs55
21
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Homework Statement


A particle initially moving at 5m/s has an acceleration of a=-.5v.
Find v(t)


Homework Equations





The Attempt at a Solution


i know the anti derivative of acceleration is -(v^2)/4
but i have no idea how to put velocity in terms of time when acceleration in terms of v is given
 
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  • #2
It's a separable differential equation:

[tex]a = \frac{dv}{dt} = -\frac{1}{2}v[/tex]

Does that help?
 
  • #3
not really...i still don't fully understand how to put v as a function of time
 
  • #4
When solving a separable differential equation, you treat the derivative (dv/dt) as a fraction. Get all the v's and dv's on one side and all the t's and dt's on the other side:
dv/dt=-1/2 v
1/v dv=-1/2 dt

Then integrate both sides and solve for v.
 
  • #5
the answer given in the notebook is "v=5 exp(-.5t)"...i don't understand :s
when i integrate i get
ln|v| = -1/2t
solving for v = e^(-.5t)
where is this 5 coming from??
 
Last edited:
  • #6
You forgot the constant in your integration. What you have is an initial condition problem: v(0)=5. So when you integrate both sides, you get
ln|v| = -1/2 t + c
v=e^(-1/2 t+c)

It makes things neater to get that constant out of the exponent, using the laws of exponents:
v=e^c*e^(-1/2 t)

Now, since c is a constant, e^c is a constant, so we can just cal lthat a new constant to make it neater:
Let k=e^c
v=k*e^(-1/2 t)

With your initial condition v(0)=5:
5=k*e^(-1/2*0)
5=k*1
k=5

So v=5e^(-1/2 t)
 
  • #7
ahh i c thanks alot!
 

1. What is the formula for calculating a particle's velocity given acceleration?

The formula for calculating a particle's velocity given acceleration is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed.

2. How does calculus help in solving for a particle's velocity given acceleration?

Calculus helps in solving for a particle's velocity given acceleration by providing a mathematical framework for understanding and analyzing the relationship between position, velocity, and acceleration. It allows us to use derivatives and integrals to find the velocity of a particle at a specific time given its acceleration.

3. Can a particle's velocity be negative when calculating with acceleration?

Yes, a particle's velocity can be negative when calculating with acceleration. This indicates that the particle is moving in the opposite direction of its initial velocity, either slowing down or changing direction.

4. What units are typically used for acceleration and velocity in this calculation?

The units used for acceleration are typically meters per second squared (m/s^2) and the units used for velocity are typically meters per second (m/s).

5. How can this calculation be applied in real-world scenarios?

This calculation can be applied in real-world scenarios in a variety of fields such as physics, engineering, and astronomy. For example, it can be used to calculate the velocity of a rocket during launch, the speed of a car accelerating on a race track, or the motion of a planet in orbit around a star.

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