The answer is B, Law of Kinetic Energy
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
As we know that KE and PE is same at a given position
so we will have as a function of position given as

also the PE is given as function of position as

now it is given that
KE = PE
now we will have




so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula

now to find the fraction of kinetic energy



now since total energy is sum of KE and PE
so fraction of PE at the same position will be

