Answer:
least to greatest- -1.25, -2/3, 2.75, 4, 9/2
Step-by-step explanation:
Here you go, hope this helps. Don’t mind the water droplets, accidentally topped my bottle over.
3.6y is the answer. If you multiply the 0.3 times the 12 it equales 3.6 and add the y because it’s a variable.
Answer:
(c) 2.034 s; (d) 8.944 cm
Step-by-step explanation:
Velocity and acceleration
s = 8cos(t) + 4sin(t)
v = -8sin(t) + 4cos(t)
a = -8cos(t) + 4sin(t)
(c) Time to first equilibrium position
The equilibrium position is where the mass hangs before it is pulled downward, that is, at s = 0.
Set s = 0 and solve for t.
![\begin{array}{rcl}0 & = & 8\cos t + 4\sin t\\-8\cos t & = & 4\sin t\\-8 & = & 4\dfrac{\sin t}{\cos t}\\\\-2 & = & \tan t\\t & = & \arctan(-2)\\& = & -1.107 \pm n\pi\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D0%20%26%20%3D%20%26%208%5Ccos%20t%20%2B%204%5Csin%20t%5C%5C-8%5Ccos%20t%20%26%20%3D%20%26%204%5Csin%20t%5C%5C-8%20%26%20%3D%20%26%204%5Cdfrac%7B%5Csin%20t%7D%7B%5Ccos%20t%7D%5C%5C%5C%5C-2%20%26%20%3D%20%26%20%5Ctan%20t%5C%5Ct%20%26%20%3D%20%26%20%5Carctan%28-2%29%5C%5C%26%20%3D%20%26%20-1.107%20%5Cpm%20n%5Cpi%5C%5C%5Cend%7Barray%7D)
If n = 1,
t = -1.107 + π = 2.034 s
(d) Distance from equilibrium position
The mass will reach its maximum distance when v = 0, that is, when it is at the peak or trough of its oscillation.
Set v = 0 and solve for t.
![\end{array}\begin{array}{rcl}0 & = &- 8\sin t + 4\cos t\\8\sin t & = & 4\cos t\\\dfrac{\sin t}{\cos t} & = & \dfrac{1}{2}\\\\\tan t & = & 0.5\\t & = & \arctan(0.5)\\& = & 0.4636 \pm n\pi\\\end{array}](https://tex.z-dn.net/?f=%5Cend%7Barray%7D%5Cbegin%7Barray%7D%7Brcl%7D0%20%26%20%3D%20%26-%208%5Csin%20t%20%2B%204%5Ccos%20t%5C%5C8%5Csin%20t%20%26%20%3D%20%26%204%5Ccos%20t%5C%5C%5Cdfrac%7B%5Csin%20t%7D%7B%5Ccos%20t%7D%20%26%20%3D%20%26%20%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5Ctan%20t%20%26%20%3D%20%26%200.5%5C%5Ct%20%26%20%3D%20%26%20%5Carctan%280.5%29%5C%5C%26%20%3D%20%26%200.4636%20%5Cpm%20n%5Cpi%5C%5C%5Cend%7Barray%7D)
If n = 0,
t = 0.4636
Then
s = 8cos(0.4636) + 4sin(0.4636) = 8×0.8944 + 4×0.4472 = 7.156 + 1.789 = 8.944 cm
The figure below shows the graphs of s and v vs t. They indicate that the mass first reaches its equilibrium position at 2.034 s, and the amplitude of its vibration is 8.944 cm.
Answer:
Step-by-step explanation:
Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown
near the earth's surface, and it moves along a curved path under the action of gravity only.
In projectile motion, the horizontal motion and the vertical motion are independent of each
other; that is, neither motion affects the other.
The horizontal component of the velocity of the object remains unchanged throughout the
motion. The vertical component of the velocity increases linearly, because the acceleration due to
gravity is constant (g=9.81 m/s2
).
= 0 cos
0 = 0 sin
To find the time of flight, determine the time the projectile takes to reach maximum height.
The time of flight is just double the maximum-height time.
Start with the equation:
= 0 + = 0 −
At maximum height,
= 0
The time to reach maximum height is
=
0
=
0 sin
=
17.5 sin 31.5°
9.81 = 0.9321
Horizontal distance, = 0
=
0 cos 0 sin
=
0
2
sin 2
2
=
17.5
2
sin(2 ∗ 31.5°)
2 ∗ 9.81 = 13.91
Answer: =
0 sin
; =
0
2
sin 2
2�