Question

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.

Explanation:

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 \(a(t)=v^\prime(t)=-\frac{1}{(1 + t)^2}\).

1. 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.

1. 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