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Carnegie Mellon University

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Problem 59 Hard Difficulty

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 $15.0 \mathrm{m} / \mathrm{s}^{2},$ as shown in Figure $\mathrm{P} 3.59$ . The coyote starts off at rest 70.0 $\mathrm{m}$ from the edge of a cliff at the instant the road-runner 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 $1.00 \times 10^{2} \mathrm{m}$ 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 15.0 $\mathrm{m} / \mathrm{s}^{2}$ )

Answer

(a) $v _ { \mathrm { x } _ { \mathrm { K } } } = 22.91 \mathrm { ms } ^ { - 1 }$
(b) $R = 359.8 \mathrm { m }$

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Video Transcript

before party weaken. First, use Delta X equaling V X initial T plus 1/2 acceleration in the ex direction T squared and so we can then see that then t. This would be the time for the coyote to travel 70 meters and so this would be equal in the square root of two Delta X, divided by the acceleration in the ex direction. And so this would be the square root of two times 70.0 meters and he has a constant acceleration of 15.0 meters per second squared. This is equaling 3.6 seconds. This would be keep this answer in mind essentially and so the minimum constant speed that the road burner must have to reach the edge in this time would be the equaling Delta X, divided by T, and so this would be equaling 70.0 meters, divided by 3.6 seconds. Therefore, the road runner must have a velocity of 22.9 meters per second. This would be our final solution for part a. Now, for part B, we know that to find we can find the initial velocity of the coyote as it goes over the edge of the cliff. This would be the initial. This would be the ex initial, given that all of the initial velocity as the coyote goes over the cliff is in the ex direction. So this would be equaling essentially the initial which we know to be zero because he's starting from rest. Plus the acceleration times multiplied by time and so this would be 15.0 meters per second squared multiplied by 3.6 seconds. And this is giving us 45.9 meters per second. And so we can then save Delta y equaling again. V. Why initial t plus 1/2 GT squared essentially Or better yet, say 1/2 times acceleration in the UAE direction. She squared and he's dropping 100 meters. We know that the initial velocity in the UAE direction is zero, and so we find that then we could say t's up to would be equal in the square root of two times Delta y divided by the acceleration of the Y direction So T's up to is gonna be equal in the square root of two multiplied by negative 100 meters. This is given that he's falling 100 meters divided by negative 9.80 meters per second squared and this is giving us 4.52 seconds. So the horizontal displacement of the coyote during this fall, Delta X would simply be v X initial multiplied by tea. Or, we can simply say er the X initial most most part by tea. And here he's keeping his acceleration. So B plus 1/2 times the acceleration in the ex direction Times t squared. My apologies because a teacher too square and so we can then say Delta X is going to be equaling 45.9 meters per second, multiplied by 4.52 seconds plus 1/2 multiplied by 15.0 meters per second squared multiplied by 4.52 seconds. Quantity squared, and we find that Delta X is equaling approximately 361 meters. This would be our final answer for part B. That is the end of the solution. Thank you for watching

Carnegie Mellon University
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