Which example provides a complete scientific description of an object in motion?
2 answers:
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Answer:
The BMX lands 5.4 m from the end of the ramp.
Explanation:
Hi there!
The position of the BMX is given by the position vector "r":
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t
x0 = initial horizontal position
v0 = initial velocity
α = jumping angle
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m
Solving the quadratic equation:
t = 1.2 s
Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:
x = x0 + v0 · t · cos α
x = 0 m + 5.9 m/s · 1.2 s · cos 40°
x = 5.4 m
The BMX lands 5.4 m from the end of the ramp.
Have a nice day!
Answer:
The depth of the well, s = 54.66 m
Given:
time, t = 3.5 s
speed of sound in air, v = 343 m/s
Solution:
By using second equation of motion for the distance traveled by the stone when dropped into a well:
Since, the stone is dropped, its initial velocity, u = 0 m/s
and acceleration is due to gravity only, the above eqn can be written as:
(1)
Now, when the sound inside the well travels back, the distance covered,s is given by:
(2)
Now, total time taken by the sound to travel:
t = t' + t''
t'' = 3.5 - t' (3)
Using eqn (2) and (3):
s = 343(3.5 - t') (4)
from eqn (1) and (4):
Solving the above quadratic eqn:
t' = 3.34 s
Now, substituting t' = 3.34 s in eqn (2)
s = 54.66 m