A body moves in a straight line so that at time \(\quantity{t}{sec.}\) its velocity \(\quantity{v}{ft./sec.}\) is given by the formula\[ v= \frac{2t-1}{4(2t+1)}. \]


  1. the displacement of the body, measured from its position when \(t=0\), at the instant when its velocity is zero;

[The numerical answer to (i) is required correct to two significant figures.]

The velocity is the rate of change of displacement. So to calculate the displacement away from its initial position at a given time, we need to integrate the velocity as a function of time.

Clearly if \(v = 0\), then \(t = 0.5\). We have \[v = \dfrac{1}{4}\left( \dfrac{2t-1}{2t+1} \right) = \dfrac{1}{4}\left( \dfrac{(2t+1)-2}{2t+1} \right) = \dfrac{1}{4}\left( 1 -\dfrac{2}{2t+1} \right).\]

Thus \[d = \dfrac{1}{4} \int v \:d t = \dfrac{1}{4}\left( t - \ln{\big\vert 2t+1\big\vert} + c \right) = \dfrac{t}{4} - \dfrac{\ln{\big\vert 2t+1\big\vert}}{4} + c'.\]

If \(d = 0\) when \(t = 0\), then \(c' = 0\), so at \(t = 0.5,\)

\[d = \dfrac{1}{8} - \dfrac{\ln{2}}{4} \approx \quantity{-0.048}{ft}.\]

Find: (ii) the acceleration of the body at this instant.

The acceleration \(a\) of a body at time \(t\) is found by differentiating the expression for its the velocity.

So here we have \(a = \dfrac{d}{dt} \dfrac{2t-1}{4(2t+1)}\). Using the quotient rule, we have

\[\begin{align*} a &= \dfrac{4(2t+1)\times 2 -(2t-1)\times 8 }{16(2t+2)^2}\\ &=\dfrac{16t+8 -16t+8}{16(2t+2)^2}\\ &=\dfrac{1}{(2t+2)^2} \end{align*}\]

Putting \(t = 0.5\) gives us \[a= \quantity{0.25}{ft./sec.^2}\]