Review question

# Can we find this body's displacement from $O$ when its velocity is $0$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8774

## Solution

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$v= \frac{2t-1}{4(2t+1)}.$

Find:

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 $a= \quantity{0.25}{ft./sec.^2}$