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

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 10.0 m/s2 (see figure). The coyote starts at rest 40.0 m from the edge of a cliff at the instant the roadrunner zips past him in the direction of the cliff.

(a) Determine the minimum constant speed the roadrunner must have to reach the cliff before the coyote.
(???) 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

(b) 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.
(???) m

(c) 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 = 10 m/s^2

a) let the speed of roadrunner is v

for roadrunrer to reach the cliff first

40/v = sqrt(2 * 40/10)

solving for v

v = 14.14 m/s

the minimum speed of road runner is 14.14 m/s

b)

time of fight of cayote ,

t = sqrt(2 *h/g)

t = sqrt(2 * 100/9.8)

t = 4.52 s

speed of cayote at the point of cliff

u = sqrt(2 * a *d)

u = sqrt(2 * 10 * 40)

u = 28.3 m/s

Now , the distance of cayote from the base is

d = 28.3 * 4.52 + 0.5 * 10 * 4.52^2

d = 230 m

cayote lands 230 m away from the edge of cliff

c)

Now , Using first equation of motion

vy = - 9.8 * 4.52

vy = -44.3 m/s

Vx = 28.3 + 10 * 4.52

vx = 73.5 m/s

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