Acceleration as a Function of Time Using Constant Power

[Chenkel]

Hello everyone! I was doing some dimensional analysis to find an equation that gives a acceleration as a function of time, using constant power. I came up with the equation $$a = k\sqrt {\frac P {mt}}$$ I differentiated velocity with respect to time in order to check my work and also checked out an online example and it turns out k is equal to $\frac 1 {\sqrt 2}$ I have two questions about the result, why is acceleration undefined when t is equal to 0, and also why does acceleration go to infinity as time approaches 0.

Any insight will be appreciated, thank you!

More details on how I checked the work:

$$Pt = {\frac 1 2}mv^2$$

$$v= \sqrt {2Pt/m}$$

$$a = \frac {dv} {dt} = \sqrt{\frac P {2mt}} $$

 

[russ_watters]

t=0 is problematic because speed and therefore kinetic energy are zero at t=0. Power must also be zero or if you set power to something finite, acceleration is undefined (tending toward infinity). So, you pick your poison.

 

[Chenkel]

t=0 is problematic because speed and therefore kinetic energy are zero at t=0. Power must also be zero or if you set power to something finite, acceleration is undefined (tending toward infinity). So, you pick your poison.

I agree that at t=0 kinetic energy will be 0, but what's problematic about having non zero power at that point ${\frac {dE} {dt}} > 0$ also I am confused why constant power can create infinite acceleration.

 

[russ_watters]

I agree that at t=0 kinetic energy will be 0, but what's problematic about having non zero power at that point ${\frac {dE} {dt}} > 0$

How do you apply or calculate power/energy at t=0? You have to define your scenario for real-life and then describe it with the math. For example, in real life an electric motor applies constant torque to a shaft at all shaft speeds. So, what's the power at t=0?

 

[russ_watters]

...a car engine with a continuously variable transmission is effectively a constant power device. How does the transmission work? As the output rpm varies, what else varies? As you go toward zero, what happens?

 

[erobz]

I think if you replace the constant power with the following relationship, you'll see what's going on.

$$ P = \vec{ \mathbf{F} } \cdot \vec{ \mathbf{v} } $$

 

[Dale]

I agree that at t=0 kinetic energy will be 0, but what's problematic about having non zero power at that point ${\frac {dE} {dt}} > 0$ also I am confused why constant power can create infinite acceleration.

Since $P=\vec f \cdot \vec v$ if you try to keep $P$ constant then as $\vec v$ goes to 0 $\vec f$ increases without bound.

 

[Chenkel]

Since $P=\vec f \cdot \vec v$ if you try to keep $P$ constant then as $\vec v$ goes to 0 $\vec f$ increases without bound.

Does this mean if I have constant power then pushing a ball forward will cause it to accelerate by a huge acceleration near t = 0 and an arbitrarily large acceleration as $t \rightarrow 0$

 

[Dale]

huge acceleration near t = 0

It is a huge acceleration near $\vec v=0$. Not necessarily near $t=0$

 

[Chenkel]

It is a huge acceleration near $\vec v=0$. Not necessarily near $t=0$

But $$a = \frac {dv} {dt} = \sqrt{\frac P {2mt}} $$ so isn't the acceleration arbitrarily large as t approaches 0? Aren't both things true when power is constant, that for small velocity, acceleration is large?

 

[Dale]

But $$a = \frac {dv} {dt} = \sqrt{\frac P {2mt}} $$ so isn't the acceleration arbitrarily large as t approaches 0? Aren't both things true when power is constant, that for small velocity, acceleration is large?

Yes, but that is only one solution with constant power. There are other solutions where power is constant but $v(0)\ne 0$. In those solutions the force increases without bound as $\vec v$ goes to 0, regardless of $t$.

When you have two things happen together you have to be careful in generalizing that you don’t accidentally generalize the wrong thing.

 

[Delta2]

@Chenkel, start from the equation $$Pt=\frac{1}{2}mv^2-\frac{1}{2}mv_0^2$$, that is assuming that we have initial velocity $v(0)=v_0\neq 0$, you will find out then that $a,v$ are finite for t=0.

 

[PeroK]

But $$a = \frac {dv} {dt} = \sqrt{\frac P {2mt}} $$ so isn't the acceleration arbitrarily large as t approaches 0? Aren't both things true when power is constant, that for small velocity, acceleration is large?

So, in practice constant power may not be physically realisable. Instead power must increase and decrease continuously from zero as speed increases from and decreases to zero.

There is a similar issue if you consider smaller and smaller masses. In your equation, the acceleration and final speed over a short initial period are unlimited IF you are genuinely capable of delivering constant power. In practice, it becomes harder to deliver power to a smaller mass. For example, you can put a lot of power into throwing something heavy (like a basketball), but not something very light, like a pea.

In fact, if you want to accelerate a pea, you are better flicking it with a finger than trying to use your big muscles, which simply cannot deliver much power to something as light as a pea.

 

[vanhees71]

Let's make the problem 1D, i.e., a car driving along a straight street. Then you have
$$m \dot{v}=F, \quad P=vF=\text{const}.$$
So you get
$$m v \dot v =P \; \Rightarrow \; \frac{m}{2} (v^2-v_0^2) =P t,$$
where in the step I have simply integrated over $t$ from $0$ to $t$, assuming an initial velocity $v_0$. So you get
$$v=\sqrt{\frac{2Pt}{m}+v_0^2}.$$
Integrating once more
$$x=x_0 + \frac{m}{3P} \left[\left(v_0^2+\frac{2 P t}{m} \right)^{3/2}-v_0^3 \right].$$

 

[Chenkel]

$$m v \dot v =P \; \Rightarrow \; \frac{m}{2} (v^2-v_0^2) =P t,$$

I agree with the equation on the left and on the right, I just don't see how the first implies the second, it seems you might be using some kinematic formula to derive the second from the first.

 

[Delta2]

I agree with the equation on the left and on the right, I just don't see how the first implies the second, it seems you might be using some kinematic formula to derive the second from the first.

Take integrals with respect to time t, in both sides of the first equality.

 

[erobz]

It more clearly realized in Leibnitz notation:

$$ m v \dot v = P $$

is

$$ m v \frac{dv}{dt} = P $$

Which implies:

$$\int m v \, dv = \int P\, dt$$

With constant mass $m$ and power $P$ and initial velocity $v_o$ and time $t_o = 0$

$$ \int m v \,dv = \frac{1}{2} m \left( v^2 - v_o^2 \right) = Pt = P\int dt $$

 

[Chenkel]

In practice, it becomes harder to deliver power to a smaller mass. For example, you can put a lot of power into throwing something heavy (like a basketball), but not something very light, like a pea.

So you're hand has a force on it $F_{hand} = (m_{hand})(a_{hand})$ and your ball has a different force but the same acceleration $F_{ball} = (m_{ball})(a_{hand})$ so this allows us to write $F_{ball} = (m_{ball})(\frac {F_{hand}} {m_{hand}})$ so we can see when the mass of the ball goes up, the force applied to the ball increases, thus increasing the power over the throwing interval which means you will impart more energy to a more massive object thrown.

 

[jbriggs444]

So you're hand has a force on it $F_{hand} = (m_{hand})(a_{hand})$ and your ball has a different force but the same acceleration $F_{ball} = (m_{ball})(a_{hand})$ so this allows us to write $F_{ball} = (m_{ball})(\frac {F_{hand}} {m_{hand}})$ so we can see when the mass of the ball goes up, the force applied to the ball increases, thus increasing the power over the throwing interval which means you will impart more energy to a more massive object thrown.

Until you try to throw your house, cannot even budge it and the delivered energy goes to zero.

 

[Chenkel]

In practice, it becomes harder to deliver power to a smaller mass. For example, you can put a lot of power into throwing something heavy (like a basketball), but not something very light, like a pea.

In fact, if you want to accelerate a pea, you are better flicking it with a finger than trying to use your big muscles, which simply cannot deliver much power to something as light as a pea.

Isn't the power (amount of work over time) that a human is able to do roughly constant? so it will take a longer time to accelerate that rock to the same velocity as the pea. Isn't it easier to deliver the power to the pea than the large mass since the power will translate into a longer distance traveled for the pea? I feel I'm missing something in my interpretation.

 

[anorlunda]

Isn't the power (amount of work over time) that a human is able to do roughly constant? so it will take a longer time to accelerate that rock to the same velocity as the pea. Isn't it easier to deliver the power to the pea than the large mass since the power will translate into a longer distance traveled for the pea? I feel I'm missing something in my interpretation.

Mechanical work is force times distance. Did you miss this?

Until you try to throw your house, cannot even budge it and the delivered energy goes to zero.

He is saying that since the distance you can move the house is zero, force times distance is zero, and work done is zero.

You are confusing effort, which is a biology question, with the physics of mechanical work.
A human uses energy just sitting there not moving, or more energy when pushing on a house, but he is not doing mechanical work in either case.

 

[jbriggs444]

Isn't the power (amount of work over time) that a human is able to do roughly constant? so it will take a longer time to accelerate that rock to the same velocity as the pea. Isn't it easier to deliver the power to the pea than the large mass since the power will translate into a longer distance traveled for the pea? I feel I'm missing something in my interpretation.

Setting the case of an immobile house to one side...

When you try to accelerate a pea, you will be expending most of your energy uselessly accelerating your hand. Only a fraction of the effort expended will be usefully applied as energy going into the pea.

So it is more difficult delivering energy to a pea (or ping-pong ball or paper wad) than to a baseball (or hammer or canoe paddle).

There is a sweet spot for human effort which professional bicyclists use. They will choose gearing so that they can use the optimal cadence for the desired velocity. A too slow cadence and you are not getting as much power as you could from the leg force. Too fast and you do not have time to build up maximum force and are wasting power flailing with the legs.

 

[PeroK]

Isn't the power (amount of work over time) that a human is able to do roughly constant?

No. It depends on the activity. How much power can you put into throwing a pea? Can you throw a pea faster than a tennis ball? Or, what about a grain of sand? Why can't you throw a grain of sand at the speed of a bullet?

 

[Dale]

Isn't the power (amount of work over time) that a human is able to do roughly constant?

Pretty much nothing is even roughly constant in biomechanics. The body is an incredibly messy machine.

 

[Delta2]

The body is an incredibly messy machine.

Yes , The two biggest mysteries of the universe are here right on this planet imho, it is the human body , and the even biggest mystery the human brain.

 

[Chenkel]

No. It depends on the activity. How much power can you put into throwing a pea? Can you throw a pea faster than a tennis ball? Or, what about a grain of sand? Why can't you throw a grain of sand at the speed of a bullet?

I believe we can see the problem more clearly when you divide the kinetic energy of the grain of sand by the kinetic energy of the hand $${KE}_{hand} = {\frac 1 2}{m_{hand}}v^2$$
$${KE}_{sand} = {\frac 1 2}{m_{sand}}v^2$$
$$ratio = \frac {{KE}_{sand}} {{KE}_{hand}} = {\frac {m_{sand}} {m_{hand}}}$$
The mass of a grain of sand is $1.28 * 10^{-5}$ and the mass of hand is $4.6 * 10^{2}$
$ratio= \frac {1.28 * 10^{-5}} {4.6 * 10^{2}} = {5.88 * 10^{-7}}$

Thus for every joule of work we put toward the hand we have to put ${5.88 * 10^{-7}}$ joules of energy toward the grain of sand, this means to impart one joule to the grain of sand we would have to impart $\frac 1 {5.88 * 10^{-7}} = {1.7 * 10^{6}}$ or $1.7$ million joules to the hand. So the percentage of power going to the grain of sand is very small.

 

[PeroK]

Thus for every joule of work we put toward the hand we have to put ${5.88 * 10^{-7}}$ joules of energy toward the grain of sand, this means to impart one joule to the grain of sand we would have to impart $\frac 1 {5.88 * 10^{-7}} = {1.7 * 10^{6}}$ or $1.7$ million joules to the hand. So the percentage of power going to the grain of sand is very small.

There's a maximum speed that we can move our hand even with nothing in it. That's a limiting factor. That's why it's possible to flick a pea with a fingertip as far as we can throw it - more or less.

 

[Chenkel]

I believe we can see the problem more clearly when you divide the kinetic energy of the grain of sand by the kinetic energy of the hand $${KE}_{hand} = {\frac 1 2}{m_{hand}}v^2$$
$${KE}_{sand} = {\frac 1 2}{m_{sand}}v^2$$
$$ratio = \frac {{KE}_{sand}} {{KE}_{hand}} = {\frac {m_{sand}} {m_{hand}}}$$
The mass of a grain of sand is $1.28 * 10^{-5}$ and the mass of hand is $4.6 * 10^{2}$
$ratio= \frac {1.28 * 10^{-5}} {4.6 * 10^{2}} = {5.88 * 10^{-7}}$

Thus for every joule of work we put toward the hand we have to put ${5.88 * 10^{-7}}$ joules of energy toward the grain of sand, this means to impart one joule to the grain of sand we would have to impart $\frac 1 {5.88 * 10^{-7}} = {1.7 * 10^{6}}$ or $1.7$ million joules to the hand. So the percentage of power going to the grain of sand is very small.

I may have made a mistake here, I believe an average hand is .46kg.