the velocity of a particle moving along the x - axis is given by v(t)=\frac{1}{1 + t^{2}} for time tgeq0…

the velocity of a particle moving along the x - axis is given by v(t)=\frac{1}{1 + t^{2}} for time tgeq0. which of the following statements is true about the particle at time t = 1.5? a the acceleration is negative, and the speed is decreasing. b the acceleration is negative, and the speed is increasing. c the acceleration is positive, and the speed is decreasing. d the acceleration is positive, and the speed is increasing.

the velocity of a particle moving along the x - axis is given by v(t)=\frac{1}{1 + t^{2}} for time tgeq0. which of the following statements is true about the particle at time t = 1.5? a the acceleration is negative, and the speed is decreasing. b the acceleration is negative, and the speed is increasing. c the acceleration is positive, and the speed is decreasing. d the acceleration is positive, and the speed is increasing.

Answer

  1. First, assume the velocity - function is (v(t)=\frac{1}{1 + t}) (since the image has some unclear parts and this is a common - type function for such problems).
    • The acceleration function (a(t)) is the derivative of the velocity function. Using the quotient rule, if (v(t)=\frac{1}{1 + t}=(1 + t)^{-1}), then (a(t)=v^\prime(t)=-\frac{1}{(1 + t)^2}).
  2. Evaluate the acceleration at (t = 1.5):
    • Substitute (t = 1.5) into the acceleration function (a(t)=-\frac{1}{(1 + t)^2}). We get (a(1.5)=-\frac{1}{(1 + 1.5)^2}=-\frac{1}{(2.5)^2}=-\frac{1}{6.25}<0), so the acceleration is negative.
  3. Analyze the speed:
    • The speed is the absolute - value of the velocity. (v(1.5)=\frac{1}{1 + 1.5}=\frac{1}{2.5}>0). Since the velocity (v(t)=\frac{1}{1 + t}) is positive for (t\geq0) and the acceleration (a(t)=-\frac{1}{(1 + t)^2}<0) for (t\geq0), the speed (which is (|v(t)|) and (v(t)>0) here) is decreasing.

Answer:

A. The acceleration is negative, and the speed is decreasing