Answer:
equilibrium position.
Explanation:
In simple harmonic motion , velocity v(t) is given by,
v(t) = -ω A sin(ωt + φ)
where
ω = angular velocity of the corresponding circular motion
A = amplitude
t = time
φ = the initial angle of the corresponding circular motion when the motion begin.
v (t) get maximized when sin value is maximized , i.e. sin =1
The particle has maximum speed when it passes through the equilibrium position.
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!
