Question

Particle Motion The position of a particle moving along a coordinate line is \(s=\sqrt{1+4 t},\) with \(s\) in meters and \(t\) in seconds. Find the particle's velocity and acceleration at \(t=6 \mathrm{sec}\) .

Short answer

The velocity of the particle at \(t=6 sec\) is \(\frac{2}{5} m/s\) and the acceleration is \(-\frac{4}{125} m/s^2\).

Step-by-step solution

Differentiate The Position Function For Velocity

Start by differentiating the position function \(s=\sqrt{1+4 t}\) with respect to time \(t\) in order to get velocity. Using the chain rule where the derivative of \(\sqrt{x}\) with respect to x is \(\frac{1}{2\sqrt{x}}\) and differentiating \(1+4t\) with respect to \(t\) gives \(4\), the derivative \(s'\) (which is the velocity, \(v\)) is: \(v=\frac{4}{2\sqrt{1+4t}}=\frac{2}{\sqrt{1+4t}}\).

Differentiate The Velocity Function For Acceleration

Now differentiate the velocity function with respect to time \(t\) to obtain acceleration. Again using the chain rule and noticing that \(\frac{1}{\sqrt{1+4t}}\) can be written as \((1+4t)^{-1/2}\), the derivative \(v'\) (which is the acceleration, \(a\)) is: \(a=-\frac{4}{(1+4t)^{3/2}}\).

Substitute Time To Get The Required Velocity And Acceleration

Finally, substitute \(t=6 sec\) in the velocity and acceleration equations. For velocity: \(v=\frac{2}{\sqrt{1+4(6)}}=\frac{2}{\sqrt{25}}=\frac{2}{5} m/s\). For acceleration: \(a=-\frac{4}{(1+4(6))^{3/2}}=-\frac{4}{125} m/s^2\).