Question

The position of a particle moving along the x-axis is given by a twice-differentiable function x(t). If x(2) < 0, x'(2) < 0, and x''(2) <0, which of the following statements must be true about the particle at time t=2?

a. The particle is moving toward the origin at a decreasing speed.
b. The particle is moving toward the origin at an increasing speed.
c. The particle is moving away from the origin at a decreasing speed.
d. The particle is moving away from the origin at an increasing speed.
e. The particle is moving away from the origin at a constant speed.

Answer 1

Answer:

c. The particle is moving away from the origin at a decreasing speed.

Step-by-step explanation:

Since x(2) which represents the particles position is less than zero, that is, x(2) < 0, it shows that the particle is at the left of the origin at t = 2.

Also, x'(2) which represents the particle's speed at t = 2 is less than zero, that is, x(2) < 0, the particles speed is negative at t = 2.

Finally, since x"(2) which represents the particle's acceleration at t = 2 is less than zero, that is, x"(2) < 0, the particle's acceleration is negative at t = 2 and thus decreasing.

So, combining these three conditions together shows that the particle is moving away from the origin at a decreasing speed at time, t = 2.

Answer 2

Since $x(2)$, which represents the subatomic particle location, is below zero, that is, $x(2) <0$, it indicates that the particles are to the left of the origin at $t = 2.$

• Therefore, $x'(2)$, which represents the particle's speed at $t = 2$, is less than zero, implying that$x(2) <0$, the subatomic particle speed at $t=2$, is negative.
• Furthermore, because $x"(2)$, which represents that particle's acceleration at $t = 2$, is less than zero, $x"(2)< 0$, the particle's acceleration at $t = 2$ is negative and consequently declining.
• Considering these three requirements reveals that at the moment $t = 2$, the particle is moving away from the nucleus at a decreasing speed.

Therefore, the answer is "Option C".

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