First slide

Rectilinear Motion

Question

A point moves in a straight line so that its displacement $' y' m$ at time $' t' \sec$ is given by $y^{2} = 1 + t^{2}$ . Its acceleration in $m / s^{2}$ at time $' t' \sec$

Easy

A

$\frac{1}{y} - \frac{t^{2}}{y^{3}}$
B

$\frac{1}{y^{2}} - \frac{t}{y^{3}}$
C

$\frac{1}{y} - \frac{t}{y^{2}}$
D

$\frac{1}{y^{2}} - \frac{t^{2}}{y^{3}}$

Solution

$\begin{array}{l} Given that y^{2} = 1 + t^{2} \\ y = {(1 + t^{2})}^{1 / 2} \\ Velocity V = \frac{d y}{d t} = \frac{1}{2} {(1 + t^{2})}^{- \frac{1}{2}} \cdot 2 t \end{array}$
$\begin{array}{l} = t {(1 + t^{2})}^{- \frac{1}{2}} \\ ∴ acceleration a = \frac{d v}{d t} = t [(- \frac{1}{2}) {(1 + t^{2})}^{- 3 / 2} \times 2 t] + {(1 + t^{2})}^{- \frac{1}{2}} \times 1 \\ = - t^{2} {(1 + t^{2})}^{- 3 / 2} + {(1 + t^{2})}^{- 1 / 2} \end{array}$
But from given data $1 + t^{2} = y^{2}$
$∴$ acceleration $a = - t^{2} {(y^{2})}^{- 3 / 2} + {(y^{2})}^{- 1 / 2}$
$a = \frac{- t^{2}}{y^{3}} + \frac{1}{y}$

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