Answer:
Step-by-step explanation:
I apologize for any plagiarism in my previous answer. Here is a new answer to the question:
(a) The motorcycle needs to cross the canyon safely by landing on the opposite side. Let's call the starting point of the motorcycle as point A and the landing point as point B. The horizontal distance between A and B is 1.00 km, and the vertical distance between them is 0.750 km.
To find the minimum constant acceleration in the x-direction required to cross the canyon safely, we can use the kinematic equation:
d = (1/2)at^2
where d is the horizontal distance between A and B, a is the constant acceleration in the x-direction, and t is the time it takes to cross the canyon.
We also know that the motorcycle is affected by gravity, which causes it to accelerate downwards with an acceleration of g = 9.81 m/s^2.
To cross the canyon safely, the motorcycle needs to land on the opposite side, so we can use the following inequality:
h <= (1/2)gt^2
where h is the vertical distance between A and B.
Substituting the given values, we have:
d = 1.00 km = 1000 m
h = 0.750 km = 750 m
Using the above equations, we can solve for the minimum acceleration required:
750 m <= (1/2) * 9.81 m/s^2 * t^2
1000 m = (1/2) * a * t^2
Solving for t in the first equation, we get:
t = sqrt((2 * h) / g) = sqrt((2 * 750 m) / 9.81 m/s^2) = 12.19 s
Substituting this value of t into the second equation, we get:
a = (2 * d) / t^2 = (2 * 1000 m) / (12.19 s)^2 = 6.96 m/s^2
Therefore, the minimum constant acceleration in the x-direction required to cross the canyon safely is 6.96 m/s^2.
(b) To find the speed at which the motorcycle lands, we can use the equation:
v_f = at
where v_f is the final velocity in the x-direction.
Substituting the given values, we get:
a = 6.96 m/s^2
t = 12.19 s
Therefore, the final velocity of the motorcycle in the x-direction is:
v_f = 6.96 m/s^2 * 12.19 s = 84.85 m/s
Therefore, the motorcycle lands with a horizontal speed of 84.85 m/s.