1. A coyote on jet skies chases a roadrunner. The Coyote\'s acceleration is a co
ID: 1532466 • Letter: 1
Question
1. A coyote on jet skies chases a roadrunner. The Coyote's acceleration is a constant 15 m/s^2. If the coyote is 120 m from the edge of a cliff at the moment the road runner passes him, headed in the same direction, (a) what would the roadrunner's constant speed have to be if he is to reach the edge of the cliff at the same time as the coyote? (b) If the cliff is 100m above the floor of the cayon, how far would the coyote land from the base of the cliff? (c) What are the components of the coyote's velocity just before he hits the ground?
2. A commuter plane flies from an airport to city A, 175 km away and in a direction 32 degrees North or East. Then it flies to city B, 150 km from city A and in a direction 22 degrees West of North. Finally it flies due West 190 km from city B to city C. Using analytic methods, find the location of city C relative to the airport where the plane started.
3. A jet liner initially moving at 300 mi/h due East enters a region where the wind is blowing at 100 mi/h in a direction 34 degrees North of East. What is the new velocity of the aircraft relative to the ground?
Explanation / Answer
here,
accelration of Coyote's , a = 15 m/s^2
s = 120 m
a)
let the road runner constant speed be v' and the time taken by both be t
s = u * t + 0.5 * a * t^2
120 = 0 + 0.5 * 15 * t^2
t = 4 s
as the disatnce travelled are equal
0 + 0.5 * a * t^2 = v' * t
v' = 0.5 * a * t
v' = 0.5 * 15 * 4 = 30 m/s
the constant speed of road runner is 30 m/s
b)
h = 100 m
the time taken to hit the ground be t1
h = 0 + 0.5 * g * t1^2
100 = 0.5 * 9.8 * t1^2
t1 = 4.52 s
the horizontal distance travelled , s1 = 0.5 * a * (t1 + t)^2 - 120
s1 = 0.5 * 15 * 8.52^2 - 120
s1= 424.4 m
c)
the vertical velocity before hitting the ground , vy = 0 + g * t1
vy = 9.8 * 4.52 = 44.3 m/s
the horizontal velocity of Coyote , vx = 0 + a * ( t1 + t)
vx = 15 * ( 4 + 4.52) = 127.8 m/s
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