A particle moving along a straight line has a velocity v m s-1, when it cleared a distance of x m. These two are connected by the relation v=√49+x. When its velocity is 1 m s-1, its acceleration is

Question

A particle moving along a straight line has a velocity v m s-1, when it cleared a distance of x m. These two are connected by the relation v=√49+x. When its velocity is 1 m s-1, its acceleration is (A) 2 m s-2 (B) 7 m s-2 (C) 1 m s-2 (D) 0.5 m s-2

Tardigrade Admin · Accepted Answer

Given: v=√49+x Squaring both sides, we get, v2 = 49 + x Differentiating both sides w.r.t. t, we get ; 2v (dv/dt)=(dx/dt) 2v (dv/dt)=v ; (dv/dt)=(1/2)=0.5 (∵ v=(dx/dt)) Acceleration, a=(dv/dt)=0.5 m s-2

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

$2 m s^{- 2}$

$7 m s^{- 2}$

$1 m s^{- 2}$

$0.5 m s^{- 2}$

Given : $v = 49 + x$
Squaring both sides, we get, $v^{2} = 49 + x$
Differentiating both sides w.r.t. $t$ , we get ; $2 v \frac{d v}{d t} = \frac{d x}{d t}$
$2 v \frac{d v}{d t} = v$ ;
$\frac{d v}{d t} = \frac{1}{2} = 0.5 (∵ v = \frac{d x}{d t})$
Acceleration, $a = \frac{d v}{d t} = 0.5 m s^{- 2}$

Q. A particle moving along a straight line has a velocity vms−1, when it cleared a distance of xm. These two are connected by the relation v=49+x​. When its velocity is 1ms−1, its acceleration is

2ms−2

7ms−2

1ms−2

0.5ms−2