An aging coyote cannot run fast enough to catch a roadrunner He purchases on eBa

Question

An aging coyote cannot run fast enough to catch a roadrunner He purchases on eBay a set of jet-powered roller skates, which provide a constant horizontal acceleration of 14.0 m/s^2 (see figure). The coyote starts at rest 50.0 m from the edge of a cliff at the instant the roadrunner zips past him in the direction of the cliff Determine the minimum constant speed the roadrunner must have to reach the cliff before the coyote. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error m/s At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote's skates remain horizontal and continue to operate while he is in flight, so his acceleration while in the air is (14.0i - 9.80j) m/s^2. The cliff is 100 m above the flat floor of a wide canyon. Determine how far from the base of the vertical cliff the coyote lands. Determine the components of the coyote's impact velocity. (Assume right is the positive x direction and up is the positive y direction.)

Explanation / Answer

Here

acceleration , a = 14 m/s^2

d1 = 50 m

a) let the acceleration is a

Using second equation of motion

50/u = sqrt(2 * s/a)

50/u = sqrt(2 * 50/14)

solving for u

u = 18.7 m/s

the constant velocity of rod runner must be 18.7 m/s

b) for the coyote

time of flight , t = sqrt(2h/g)

t = sqrt(2 * 100/9.8)

t = 4.52 s

Now, for the velocity of coyote at the cliff , u = 14 * 50/18.7

u = 37.4 m/s

distance from the base of cliff = 37.4 * 4.52 + 0.5 * 14 * 4.52^2

distance from the base of cliff = 312.2 m

c) for the component of impact velocity

vx = 37.4 + 14 * 4.52 = 100.68 m/s

vy = -9.8 * 4.52 = - -44.3 m/s

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