Question

A particle moving along a straight line has a velocity $vms^{- 2}$, when it cleared a distance of x m. These two are connected by the relation $v = \sqrt{49 + x.}$ When its velocity is $1ms^{- 1},$ its acceleration is

• A. $2ms^{- 2}$
• B. $7ms^{- 2}$
• C. $1ms^{- 2}$
• D. $0.5ms^{- 2}$

Answer 1

Answer: (D)

$0.5ms^{- 2}$

Answer 2

Given : $v = \sqrt{49 + x}$

Squaring both sides, we get $v^{2} = 49 + x$

Differentiating both sides w.r.t. t, we get

$2v\frac{dv}{dt} = \frac{dx}{dt}$

$2v\frac{dv}{dt} = v$ $\left( \because v = \frac{dx}{dt} \right)$

$\frac{dv}{dt} = \frac{1}{2} = 0.5$

Accelerations, a $= \frac{dv}{dt} = 0.5ms^{- 2}$

A particle moves with a constant accelerations whose magnitude is $0.5ms^{- 2}.$