The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of power roller skates, which provide a constant horizontal acceleration of 15 m/s2, as shown in the figure below. The coyote starts off at rest 70 m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff.
(a) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have to reach the cliff before the coyote.
m/s
(b) If the cliff is 100 m above the base of a canyon, find where the coyote lands in the canyon. (Assume that his skates are still in operation when he is in "flight" and that his horizontal component of acceleration remains constant at 15 m/s2.)
m from the base
To find the minimum speed the roadrunner must have to reach the cliff before the coyote, we can use the equations of motion.
(a) First, let's analyze the motion of the coyote. The horizontal motion is not affected by the vertical motion since there is no horizontal force acting on the coyote. Therefore, we can treat the horizontal and vertical motions separately.
In the horizontal direction, the coyote has a constant horizontal acceleration of 15 m/s^2. The initial velocity (u) is 0 m/s since the coyote starts from rest. The distance covered (s) is 70 m.
We can use the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values, we get:
70 = 0*t + (1/2)*(15)*(t^2)
70 = 7.5t^2
Rearranging the equation, we have:
t^2 = 70/7.5
t^2 = 9.33
t ≈ 3.05 seconds
Now, let's analyze the motion of the roadrunner. The roadrunner moves with a constant speed, let's call it v. The distance covered is 70 m, and since the speed is constant, we can use the equation: s = vt.
Plugging in the values, we get:
70 = v*t
Since we already found t ≈ 3.05 seconds, we can substitute it into the equation:
70 = v*3.05
Now, we can solve for v:
v = 70/3.05
v ≈ 22.95 m/s
Therefore, the minimum speed the roadrunner must have to reach the cliff before the coyote is approximately 22.95 m/s.
(b) To determine where the coyote lands in the canyon, we need to find the time it takes for the coyote to reach the cliff. For that, we can use the equation of motion: s = ut + (1/2)at^2.
In the vertical direction, the acceleration is due to gravity and is approximately -9.8 m/s^2 (taking downward as the positive direction). The initial velocity is 0 m/s since the coyote starts from rest. The distance covered is the height of the cliff, which is given as 100 m.
Plugging in the values, we get:
100 = 0*t + (1/2)*(-9.8)*(t^2)
100 = -4.9t^2
Rearranging the equation, we have:
t^2 = -100/(-4.9)
t^2 = 20.41
t ≈ 4.52 seconds
Now that we know the time it takes for the coyote to reach the cliff, we can use this time to find how far it travels horizontally. Since the horizontal acceleration is still 15 m/s^2, and the initial velocity is 0 m/s, we can use the equation: s = ut + (1/2)at^2.
Plugging in the values, we get:
s = 0*t + (1/2)*(15)*(t^2)
s = 7.5t^2
Substituting t ≈ 4.52 seconds, we have:
s = 7.5*(4.52)^2
s ≈ 154.31 meters
Therefore, the coyote lands in the canyon approximately 154.31 meters from the base.