# The determined Wile E. Coyote is out once

The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of power roller skates, which provide a constant horizontal acceleration of 15 m/s2, as shown in Figure P3.73. The coyote starts off at rest 70 m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff. (a) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have to reach the cliff before the coyote. (b) If the cliff is 100 m above the base of a canyon, find where the coyote lands in the canyon. (Assume his skates are still in operation when he is in “flight” and that his horizontal component of acceleration remains constant at 15 m/s2.)
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Vasquez

(a) Derive the expression for the minimum time required for the coyote to travel a distance of $\mathrm{\Delta }x$ from the cliff.

The distance travelled by the coyote is,

$\mathrm{\Delta }x={v}_{0x}t+\frac{1}{2}{a}_{x}{t}^{2}$

Here ${v}_{0x}$ is the initial horizontal velocity and ${a}_{x}$ is the horizontal acceleration of the coyote, and $t$ is the time to cover the distance $\mathrm{\Delta }x$.

The initial velocity og the coyote is zero.

Substitute  for ${v}_{0x}$ in the equation $\mathrm{\Delta }x={v}_{0x}t+\frac{1}{2}{a}_{x}{t}^{2}$ and solve for $t$.

$t=\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}$

The roadrunner is moving with a constant speed. Hence, the minimum speed to reach the cliff is as follows.

${v}_{min}=\frac{\mathrm{\Delta }x}{t}$

Substitute $\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}$ for $t$.

${v}_{min}=\frac{\mathrm{\Delta }x}{\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}}$

$=\sqrt{\frac{{a}_{x}\mathrm{\Delta }x}{2}}$

Substitute  for ${a}_{x}$ and for $\mathrm{\Delta }x$

Hence, the required minimum speed of the roadrunner is

b) The initial velocity of the coyote when it goes over the cliff is equal to the horizontal velocity.

${v}_{0x}={a}_{x}t$

Substitute $\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}$ for $t$.

${v}_{0x}={a}_{x}\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}$

$=\sqrt{2{a}_{x}\mathrm{\Delta }x}$

The time taken to cover the vertical height of the cliff is,

${t}^{\prime }=\sqrt{\frac{2\mathrm{\Delta }y}{{a}_{y}}}$

Here $\mathrm{\Delta }y$ is vertical heisght of the cliff.

Therefore, the horizontal displacement of the fall of the coyote is,

$\mathrm{\Delta }x={v}_{0x}t+\frac{1}{2}{a}_{x}{t}^{2}$

Substitute  $=\sqrt{2{a}_{x}\mathrm{\Delta }x}$ for v_{0x} and $\sqrt{\frac{2\mathrm{\Delta }y}{{a}_{y}}}$ for ${t}^{\prime }$.

$\mathrm{\Delta }x=\sqrt{2{a}_{x}\mathrm{\Delta }x}\left(\sqrt{\frac{2\mathrm{\Delta }y}{{a}_{y}}}\right)+\frac{1}{2}{a}_{x}\left(\sqrt{\frac{2\mathrm{\Delta }y}{{a}_{y}}}{\right)}^{2}$

Substitute  for ${a}_{x}$ and for $\mathrm{\Delta }x$, for $\mathrm{\Delta }y$, and for ${a}_{y}$

Hence, the reauired dispplacement is