Wile E Coyote, armed with ACME^TM-brand rocket- propelled rollerblades, is at it

Question

Wile E Coyote, armed with ACME^TM-brand rocket- propelled rollerblades, is at it again trying to catch his nemesis Road Runner. Suppose the ground velocities, in hundred miles per hour, of the twosome are described by the equations below: Wile E Coyote: v' = 200 - 8v^2, Road Runner: v' = 160 - 5v^2. Without solving the 2 equations, determine whether Wile E could (finally) catch Road Runner. What is his maximum velocity (i.e. his limiting velocity)? Solve Wile E Coyote's equation to find his velocity as a function of time. You may leave your answer in implicit form.

Explanation / Answer

The limiting speed (not velocity -- velocity is a vector quantity, and we are not given information on the direction of motion in this question) occurs when the frictional force due to air resistance is equal and opposite to the "running force" the animal is capable of generating, so that the net force is zero.

From Newton's second law, |F| = m*|a|,

so if |F| = 0, (and the mass, m is nonzero), the magnitude of the acceleration, |a| = 0.

The acceleration is equal to dv/dt, so when we are looking for the limiting velocities, we are looking for the velocities for which |a| = |dv/dt| = 0

For the Coyote, the limiting velocity is the value of v such that: v' = 0 = 200 - 8v^2 v max_WEC= +sqrt(200/8) = sqrt(25) = 5 (hundred mph)

Similarly, for the Road Runner: 0 = 160- 5v^2 vmax_RR = sqrt(160/5) = sqrt(32) = 4*sqrt(2) ~= 5.66 (hundred mph). Tell your teacher (s)he writes poor questions. You cannot answer whether Wile E can catch the Road Runner unless you are told the initial positions and speeds (I'll grant that the velocities, i.e., the directions of motion, are in the same direction and along a straight line).

Both creatures have speeds that are described by differential equations of the form:

dv/dt = a - b*v^2 dv/dt = b*(a/b - v^2)

This is a separable equation:

dv/(a/b - v^2) = b dt

The integral of the left hand side is just one of those things you have to memorize (or look up in a table of integrals): sqrt(b/a) * arctanh(v*sqrt(b/a) = b*t + c

where c is the constant of integration.

arctanh(v*sqrt(b/a) = sqrt(a/b)*(b*t + c)

arctanh(v*sqrt(b/a) = sqrt(a*b)*t + C

where C = c*sqrt(a/b) is just another way of writing the constant.

v*sqrt(b/a) = tanh(sqrt(a*b)*t + C)

v(t) = sqrt(a/b) * tanh(sqrt(a*b)*t + C)

This is the explicit solution for v(t) Without knowing the initial speeds of the animals, you cannot solve for the integration constant, C, and therefore cannot obtain a particular solution.

If we assume that the animals start from rest, then v(0) = 0, so:

v(0) = 0 = sqrt(a/b)*(tanh(C))

=> C = 0

So WEC's speed as a function of time (assuming he starts from rest) is given by:

v_WEC(t) = sqrt(200/8) * tanh(sqrt(1600)*t) = 5*tanh(40*t)

and the Road Runner's speed (assuming he starts from rest) is given by:

v_RR(t) = sqrt(160/5)*tanh(sqrt(800)*t) = 4*sqrt(2)*tanh(20*sqrt(2)*t)

Once again, I encourage you to tell your teacher that this is a poorly posed problem. You cannot tell whether the coyote can catch the road runner unless you know their initial positions (relative to one another), their initial speeds, and the directions of their velocities.

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