When 7000 hp Just Isn't Enough

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Photo by Lt. Col. Derik Crotts, courtesy of U.S. Army

Top-fuel dragsters like the one shown above accelerate from rest at a tremendous rate. They race on a straight 1/4 mile long track. From a standing start, they complete the quarter mile in about 4.5 seconds and reach a speed of about 330 mph by the finish line.

Part A

Show that the statistics given in the problem introduction are inconsistent at the 10% level with the assumption that the dragster produces constant acceleration as it moves down the track.

Solution

System: Dragster as point particle.

Interactions:

Model:

Approach:

We are asked to prove that the model we are examining does not fit the data. To do so, we look for a contradiction. In this case, we have several ways to find the acceleration, since we have a large number of givens. Choosing our coordinates such that the race begins at x_i = 0 and proceeds in the positive x direction, we have:

Givens

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\begin

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[ t_

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= 0][t_

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=\mbox

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][ x_

= 0][x_

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= \mbox

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][v_

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= 0][v_

= \mbox

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]\end

We now have three separate paths to find the acceleration a. First, we can use:

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\begin

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[ v_

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= v_

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+ at ]\end

to give:

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\begin

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[ a = \frac{v_{f}}

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= \mbox

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^

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]\end

Next, we can use:

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\begin

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[ x_

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= x_

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t + \frac

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= \frac

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]\end

to give:

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[ a = \frac{2x_{f}}{t^{2}} = \mbox

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^

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]\end

Finally, we can use:

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[ v_

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^

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^

+ 2a(x_

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) ]\end

to give:

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[ a = \frac{v_

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^{2}}{2x_{f}} = \mbox

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^

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]\end

These accelerations differ from each other by more than 10%, so we conclude that the model is inconsistent at the 10% level.

Part B

Using the fact that a top-fuel dragster will typically weigh 2200 lbs (equivalent to a mass of 1000 kg), show that the assumption that the dragster moves down the track with a constant value for the power delivered to the car in the form of kinetic energy is consistent at (approximately) the 10% level.

Solution

System: Dragster as point particle.

Interactions:

Model:

Approach:

To show that the model is consistent, we need two ways to find the power that rely on different subsets of the given information. The most straightforward way to find the power is to use the velocity and the time, since the Law of Change for our model is:

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= K_

+ P^

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\Delta t = E_

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]\end

Thus, we can write:

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\begin

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[ P^

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= \frac{\frac

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mv_

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^{2}}

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= 2.42\times 10^

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\mbox

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= \mbox

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]\end

We have so far used the time and the velocity information, but we have not yet used the fact that the race is 1/4 mile long. To bring in distance information, we must perform an integral. From the definition of power, we know that we can write:

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\begin

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v ]\end

where v is the speed of the dragster. This equation can be integrated using a useful trick.

The trick we are about to use is one that might serve you well in other problems.

If we use the fact that the non-conservative force is equal to the net force on the dragster in this case, we can write:

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[ P^

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= mav = m\frac

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v ]\end

we can now use the chain rule to write:

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= \frac

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= \frac

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v]\end

This gives us the integrable equation:

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\begin

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[ P^

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dx = mv^

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dv ]\end

giving:

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[ P^

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(x_

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m(v_

^

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v_
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^

) ]\end

This equation might be considered to express "the kinematics of constant power" as opposed to the kinematics of constant force that we have spent so much time learning.

We can now express the power in terms of the distance traveled and the velocity gained:

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= \frac{mv_

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^{{3}}{3x_{f}} = 2.7\times 10}

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\mbox

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= \mbox

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]\end

which is consistent with our previous answer at about the 10% level.

It is usually true that when attempting to maximize acceleration, the motion is power-limited. This is why sports cars laud their horsepower ratings. Maximal braking however, is usually better described by motion with constant acceleration.

The upshot of this is that although these dragsters' engines deliver around 3500 hp (over ten times the rated horsepower of a production sports car) the car's performance is still effectively power limited. The capacity of friction to produce acceleration is not the limiting factor. (If you considered the values obtained for the average acceleration in Part A, this should come as something of a surprise.) It is worth noting that although we have calculated the power delivered to the kinetic energy of the car, the rated horsepower of engines is found using a different method. Due to losses in the drive train, the rated horsepower of an engine is usually about double the delivered horsepower. This is true in the case of top fuel dragsters, whose engines are rated at over 7000 hp.

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Child pages

When 7000 hp Just Isn't Enough

Part A

Solution

Part B

Solution