Question

A particle starts at \(x=0\) and moves along the \(x\) -axis with velocity \(v(t)=2 t+1\) for time \(t \geq 0 .\) Where is the particle at \(t=4 ?\)

Short answer

At \( t = 4 \) seconds, the particle is at position \( x = 12 \) units.

Step-by-step solution

Define the Velocity Function

The velocity function is already provided: \( v(t)=2t+1 \).

Integrate the Velocity function over the desired interval

Find the displacement of the particle by integrating the velocity from \( t=0 \) to \( t=4 \). This will be done using the fundamental theorem of calculus, which states that the integral of a function over an interval is equal to the antiderivative of the function evaluated at the upper limit of the interval minus the antiderivative of the function evaluated at the lower limit of the interval. Mathematically: \[ x = \int_0^4 v(t)dt = \int_0^4 (2t+1)dt \]

Solve the Definite Integral

Calculate the integral: \[ x = [\frac{2}{2}t^2+t]_0^4 = (\frac{1}{2}*4^2+4) - (\frac{1}{2}*0^2+0) = 12 \]