Physics kinematics calculus question

[shiv_99]

Homework Statement

Initial acceleration of a particle moving in a straight line is a° and initial velocity is zero.
The acceleration reduces continuously to half in every t° seconds. What will be the terminal velocity(v)?

Homework Equations

[itex]\int[/itex]adt=[itex]\int[/itex]dv

The Attempt at a Solution

[itex]^{∞}_{0}[/itex][itex]\int[/itex]adt=[itex]^{v}_{0}[/itex][itex]\int[/itex]dv
(velocity will become terminal only when accelaration becomes 0 and since acceleration is becoming half after every time interval, it will become zero at infinity)
But to use this equation I need to find an expression for acceleration in terms of time. How should i do it? Should i use geometric progression?

 

[Simon Bridge]

a(t) is a decaying exponential - the time to reduce by half is called the "half life".
You can derive the formula if you like ... start with da/dt=λa knowing that a(t°)=a°/2

You should find that the velocity-time graph is asymptotic to some v_max.

 

[Infinitum]

Hi shiv99, welcome to PF

Homework Statement

Initial acceleration of a particle moving in a straight line is a° and initial velocity is zero.
The acceleration reduces continuously to half in every t° seconds. What will be the terminal velocity(v)?

Homework Equations

[itex]\int[/itex]adt=[itex]\int[/itex]dv

The Attempt at a Solution

[itex]^{∞}_{0}[/itex][itex]\int[/itex]adt=[itex]^{v}_{0}[/itex][itex]\int[/itex]dv
(velocity will become terminal only when accelaration becomes 0 and since acceleration is becoming half after every time interval, it will become zero at infinity)
But to use this equation I need to find an expression for acceleration in terms of time. How should i do it? Should i use geometric progression?

The final velocity of the particle for the first t seconds is given as [itex]v_1=at[/itex], for the next t seconds is, [itex]v_2= at+at/2[/itex] and so on.

So yes, you can apply the infinite geometric sum for these.

 

[Curious3141]

Hi shiv99, welcome to PF



The final velocity of the particle for the first t seconds is given as [itex]v_1=at[/itex], for the next t seconds is, [itex]v_2= at+at/2[/itex] and so on.

So yes, you can apply the infinite geometric sum for these.

This would only be correct if the particle's acceleration was halving *discontinuously* every t₀ seconds.

The correct approach in this case is to set up a differential equation for a, i.e. [itex]\frac{da}{dt} = -{\lambda}a[/itex] and solve it. Basically, this is the same equation as in exponential radioactive decay.

After integrating (use the conditions given to determine the appropriate bounds), you'll get a closed form expression for a(t).

Now [itex]v(t) = \int_0^t a(t)dt[/itex].

Solve for v(t) and find [itex]\lim_{t \rightarrow \infty} v(t)[/itex] to determine the terminal velocity. The answer should have [itex]\ln 2[/itex] in it.

 

[Infinitum]

This would only be correct if the particle's acceleration was halving *discontinuously* every t₀ seconds.

Ahh, my bad.

 

[Simon Bridge]

@Infinitum: you had me panicking for a bit there :)

 

[shiv_99]

This would only be correct if the particle's acceleration was halving *discontinuously* every t₀ seconds.

The correct approach in this case is to set up a differential equation for a, i.e. [itex]\frac{da}{dt} = -{\lambda}a[/itex] and solve it. Basically, this is the same equation as in exponential radioactive decay.

After integrating (use the conditions given to determine the appropriate bounds), you'll get a closed form expression for a(t).

Now [itex]v(t) = \int_0^t a(t)dt[/itex].

Solve for v(t) and find [itex]\lim_{t \rightarrow \infty} v(t)[/itex] to determine the terminal velocity. The answer should have [itex]\ln 2[/itex] in it.

Here what will be the value of lambda for this ques. Plz help because this(exponential radioactive decay)is a new concept for me

 

[Simon Bridge]

Lambda is a dummy constant - you find out what it is after you solve the differential equation by using the fact that you know one of the solutions, vis: a(t°)=a°/2 ... you'll also get a constant of integration which you find from a(0).

 

[shiv_99]

But how do you use it find a(t)

 

[Simon Bridge]

You find a(t) by solving the differential equation provided.

One last time: a(t) is the solution to: [itex]\frac{da}{dt} = \lambda a[/itex] knowing that [itex]a(t=0)=a_0[/itex] and [itex]a(t=T_{1/2})=a_0/2[/itex]

Or you can google "exponential decay" or "radioactive decay" and use the solution provided.

 

[Curious3141]

Here what will be the value of lambda for this ques. Plz help because this(exponential radioactive decay)is a new concept for me

Simon has already covered the essentials. You need to express [itex]\lambda[/itex] in terms of [itex]t_0[/itex] by solving the differential equation, then using the "half-life" condition stated.

 

[Simon Bridge]

I'm allowing that Shiv may not know how to solve differential equations ... though just looking up the solution and adapting it to this specific case is also valid. It wouldn't be the first time someone was doing a problem without the broader knowledgebase the rest of us take for granted. Nothing wrong with that as such - it shows the ambition and courage of the self taught.