The determined coyote is out once more in pursuit of the elusive roadrunner. The
ID: 1428206 • Letter: T
Question
The determined coyote is out once more in pursuit of the elusive roadrunner. The coyote wears a pair of Acme jet-powered roller skates, which provide a constant horizontal acceleration of 10.0 m/s^2 (see figure). The coyote starts at rest 60.0 m from the brink of a cliff at the instant the roadrunner zips past him in the direction of the cliff. The roadrunner moves with constant speed. Determine the minimum speed he must have so as to reach the cliff before the coyote. 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 that his acceleration while in the air is (10.0i - 9.80j) m/s^2. The cliff is 100 m above the flat floor of a wide canyon. Determine where the coyote lands in the canyon. 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)
for coyote :
distance from cliff = d = 60 m
acceleration = a = 10 m/s2
time taken = t
initial speed = Vi = 0 m/s
Using the equation
d = Vi t + (0.5) a t2
60 = 0 x t + (0.5) (10) t2
t = 3.5 sec
Final velocity at the cliff is given as
Vf2 = Vi2 + 2 a d
Vf2 = 02 + 2 (10) (60)
Vf = 34.6 m/s
for the roadrunner :
V = constant speed
t = time = 3.5 sec
d = distance = 60 m
V = d/t = 60/3.5 = 17.14 m/s
so minimum speed = 17.14 m/s
b)
consider the motion along Y-direction :
Y = height = - 100 m
ay = acceleration = - 9.8 m/s2
t = time
Viy = 0
Using the equation
Y = Viy t + (0.5) ay t2
- 100 = 0 t + (0.5) (-9.8) t2
t = 4.52 sec
consider the motion along X-direction :
X = horizontal distance from base of cliff
ax = acceleration = 10 m/s2
t = time
Vix = 34.6 m/s
Using the equation
X = Vix t + (0.5) ax t2
X = 34.6 (4.52) + (0.5) (10)(4.52)2
X = 258.54 m
C)
Along the X-direction :
final velocity is given as
Vfx = Vix + ax t = 34.6 + (10) (4.52) = 79.8 m/s
Along the Y-direction :
final velocity is given as
Vfy = Viy + ay t = 0 + (9.8) (4.52) = 44.3 m/s