Solving the Unstoppable Coyote's Speed Challenge

[bkoz316]

Constant Speed?

The determined Wile E. Coyote is out once more to try to capture the elusive Road Runner. The coyote wears a pair of acme power roller skates, which provide a constant horizontal accleration of 15 m/s^2. The coyote starts off at rest 70m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff.

(a) If the road runner moves with a constant speed, find the maximum speed the road runner must have in order to reach the cliff before the coyote.

(b) If the cliff is 100m above the base of the canyon, find where the coyote lands in the canyon. (Assume that his skates are still in operation when he is in flight and that his horizontal component of acceleration remains constant at 15 m/s^2).





There were 10 parts to this problem and I am stuck on how to do these two.
Thanks soooooo much for the help!

 

[Yuravian]

The determined Wile E. Coyote is out once more to try to capture the elusive Road Runner. The coyote wears a pair of acme power roller skates, which provide a constant horizontal accleration of 15 m/s^2. The coyote starts off at rest 70m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff.

(a) If the road runner moves with a constant speed, find the maximum speed the road runner must have in order to reach the cliff before the coyote.

(b) If the cliff is 100m above the base of the canyon, find where the coyote lands in the canyon. (Assume that his skates are still in operation when he is in flight and that his horizontal component of acceleration remains constant at 15 m/s^2).





There were 10 parts to this problem and I am stuck on how to do these two.
Thanks soooooo much for the help!

For this type of problem, remember that 'constant speed' means a=0. So for part a, you want to find how long it takes Wile E. to get to the edge of the cliff, and then find a velocity for which the roadrunner will take the same time t to get to the edge as Wile E. Since anything faster than this means he arrives before Wile E., your answer would probably be 'greater than v meters/second'

 

[ogb p]

I would approach this problem by first stating the known information and then using the principles of physics to solve for the unknown variables.

(a) The known variables in this problem are: the constant horizontal acceleration of the coyote (15 m/s^2), the initial distance between the coyote and the cliff (70m), and the height of the cliff (100m). The unknown variable is the maximum speed of the road runner.

To solve this, we can use the equation: distance = initial velocity x time + 1/2 x acceleration x time^2. In this case, the initial velocity is 0 since the coyote starts at rest. The distance is 70m and the acceleration is 15 m/s^2. We can rearrange the equation to solve for time: time = square root of (2 x distance/acceleration).

Plugging in the values, we get: time = square root of (2 x 70m/15 m/s^2) = 4.71 seconds. This is the time it takes for the coyote to reach the edge of the cliff. Since we know that the road runner must reach the cliff before the coyote, its maximum speed must be greater than or equal to the distance divided by time: speed = 70m/4.71s = 14.87 m/s. Therefore, the road runner must have a speed of at least 14.87 m/s to reach the cliff before the coyote.

(b) To find where the coyote lands in the canyon, we need to find the distance it travels horizontally before reaching the edge of the cliff. This can be calculated using the equation: distance = initial velocity x time + 1/2 x acceleration x time^2. In this case, the initial velocity is 0 and the acceleration is still 15 m/s^2. The time is the same as calculated in part (a): 4.71 seconds. Plugging in the values, we get: distance = 0 x 4.71s + 1/2 x 15 m/s^2 x (4.71s)^2 = 176.6m. This is the distance the coyote travels horizontally before reaching the cliff.

Since the cliff is 100m high, the coyote will continue to travel horizontally for 76.6m after reaching the edge of the cliff before falling down. Therefore, the