Answer:
a)
, b)
,
, 
Explanation:
a) The system mass-spring is well described by the following equation of equilibrium:

After some handling in physical and mathematical definition, the following non-homogeneous second-order linear differential equation:
The solution of this equation is:

The velocity function is:

Initial conditions are:

Equations at
are:

The spring constant is:


After some algebraic handling, amplitude and phase angle are found:


The position can be described by this function:

b) The period of the motion is:


The amplitude is:

The phase of the motion is:

Answer:
The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
Explanation:
Given;
the speed of the slower runner, u₁ = 11.8 km/hr
the speed of the fastest runner, u₂ = 15 km/hr
distance, d = 8 km
The time when the fastest runner finishes the race is given by;

The distance covered by the slower runner at this time is given by;
d₁ = u₁ x 0.533 hr
d₁ = 11.8 km/hr x 0.533 hr
d₁ = 6.29 km
Additional distance (x) the slower runner need to finish is given by;
6.29 km + x = 8km
x = 8 k m - 6.29 km
x = 1.71 km
Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
Answer:
The minimum speed the car must have at the top of the loop to not fall = 35 m/s
Explanation:
Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)
mv²/r = mg
v² = gr = 9.8 × 25 = 245
v = 15.65 m/s
But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop
Change in kinetic energy = potential energy at the top
Change in kinetic energy = (mv₂² - mv₁²)/2
v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s
v₂ = minimum velocity the car must have at the top of the loop to not fall
And potential energy at the top of the loop = mgh (where h = the diameter of the loop)
(mv₂² - mv₁²)/2 = mgh
(v₂² - v₁²) = 2gh
(v₂² - (15.65)²) = 2×9.8×50
v₂² - 245 = 980
v₂² = 1225
v₂ = 35 m/s
Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s
To keep cars safe...............