A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π
b. vy = 4π cos (4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]
d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax
i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
Answer:
A. a set of mathematically topics that are relevant to introductory physics.
Explanation:
The physics primer is not defined as the online comprehensive mathematics textbooks. It is the set of topics of mathematics which gives students trouble and remember.
Therefore, it is defined as the process of physics problem solving. So, mathematically skills are covered in physics course as a primer related success.
Therefore, it is a set of topics of mathematics that are relevent to introductory physics.
A. a child gently swinging on a swing at small angles all the time
Answer:
a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared
b) The force of the SUV's bumper on the truck's bumper is 18000 newtons
Explanation:
a) By Newton's second law we can find the relation between force and acceleration of the SUV:

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.
Answer:
The answer is not correct.
Explanation:
Stu's answer is not correct, the equation to use is known as the law of ohm. In which the voltage is defined as the product of the current by the resistance, then we will see this equation.
![V = I*R\\where:\\I = current [amp]\\R = resistance [ohm]\\V = voltage [volts]\\](https://tex.z-dn.net/?f=V%20%3D%20I%2AR%5C%5Cwhere%3A%5C%5CI%20%3D%20current%20%5Bamp%5D%5C%5CR%20%3D%20resistance%20%5Bohm%5D%5C%5CV%20%3D%20voltage%20%5Bvolts%5D%5C%5C)
In order to find resistance, this term is found multiplying the current on the right side of the equation, therefore the current will be divided on the left side of the equation.
![R=\frac{V}{I} \\replacing:\\R=\frac{4}{0.5} \\R=8[ohms]](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BV%7D%7BI%7D%20%5C%5Creplacing%3A%5C%5CR%3D%5Cfrac%7B4%7D%7B0.5%7D%20%5C%5CR%3D8%5Bohms%5D)
That is the reason that the result found by Stu is not correct.