<span>f(x) = 5.05*sin(x*pi/12) + 5.15
First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.
So our function at this point looks like
f(x) = 5.05*sin(x*pi/12)
But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
f(x) = 5.05*sin(x*pi/12) + 5.15
The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
f(x) = 5.05*sin(x*pi/12 + C) + 5.15
where
C = Phase correction offset.
But we don't need it for this problem, so the answer is:
f(x) = 5.05*sin(x*pi/12) + 5.15
Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>
Answer:
a) 
b) the motorcycle travels 155 m
Explanation:
Let 
, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):
where:
is the speed of the motorcycle at time 2
is the velocity of the car (constant)
is the velocity of the car and the motorcycle at time 1
d is the distance between the car and the motorcycle at time 1
x is the distance traveled by the car between time 1 and time 2
Solving the system of equations:
![\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dcar%26motorcycle%5C%5Cx%3Dv_0%5CDelta%7Bt%7D%26x%2Bd%3D%28%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%7D%29%20%5CDelta%7Bt%7D%5Cend%7Barray%7D%5Cright%5D)
For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:
Answer:force equals to rate of change of momentum
Explanation:
F=force
t=time
m=mass
v=final velocity
u=initial velocity
(mv-mu)/t=rate of change of momentum
Force=rate of change of momentum
F=(mv-mu)/t
Answer:
-2.79 m/s²
Explanation:
Given:
v₀ = 20 m/s
v = 11 m/s
Δx = 50 m
Find: a
v² = v₀² + 2aΔx
(11 m/s)² = (20 m/s)² + 2a (50 m)
a = -2.79 m/s²
Round as needed.
Answer:
1.d
2.d
3.c
4.c
Okay, d & d, c & c. Iknow what you're thinking, I guessed. Nope! I just took a good look and chosed what I thought was the best answer
Explanation:
Have a great day! Byeeee
