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 20.0m/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

a_x = 20m/s^2
S = 40 m
u = 0
a) Speed of coyote at the edge of cliff :
S = ut + 0.5 at^2
40 = 0.5 *20 *t^2
t = 2 s
v_edge = u +at = 0 + 20*2 = 40 m/s

b)
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
(20.0i 9.80j) m/s^2.
Height ,h = 100 m
a = (20.0i 9.80j) m/s^2
time of flight ,T
h = 0 + 9.8 t^2
100 = 9.8 t^2
t = 3.19 s
to calculate range we, take the horizontal components of a.
at edge v_edge = 40 m/s
at landing site let velocity be v_land.
t = 3.19s
let ranvge be R , then
R = v_edge *3.19 + 0.5 *20*3.19^2 = 229.36 m from the base of cliff

c) impact velocity:
v_x = v_edge +20*3.19 = 103.8 m/s
v_y = 0 + 9.8 *3.19 = 31.26 m/s

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