Velocity (v) versus displacement (x) plot of a body moving along a straight line is as shown in the graph. The corresponding plot of acceleration (a) as a function of displacement (x) is

Question

Velocity (v) versus displacement (x) plot of a body moving along a straight line is as shown in the graph. The corresponding plot of acceleration (a) as a function of displacement (x) is Question

(A)

(B)

(C)

(D)

Tardigrade Admin · Accepted Answer

The given graph v=Kx .....(i) Solution

(d v/d t)=K . v (from the equation (i)) (d v/d t)=K.(K x)=K2x a=K2x and v= -Kx+v0 (d v/d t)= -K.v= -K(- K x + v0) a=K2x-Kv0 .......(ii) So, according to the equation (ii). Solution

Q. Velocity $(v)$ versus displacement $(x)$ plot of a body moving along a straight line is as shown in the graph. The corresponding plot of acceleration $(a)$ as a function of displacement $(x)$ is

The given graph
$v = K x$ .....(i)

$\frac{d v}{d t} = K . v$ (from the equation (i))
$\frac{d v}{d t} = K . (K x) = K^{2} x$
$a = K^{2} x$
and $v = - K x + v_{0}$
$\frac{d v}{d t} = - K . v = - K (- K x + v_{0})$
$a = K^{2} x - K v_{0}$ .......(ii)
So, according to the equation (ii).

Q. Velocity (v) versus displacement (x) plot of a body moving along a straight line is as shown in the graph. The corresponding plot of acceleration (a) as a function of displacement (x) is

The given graph v=Kx .....(i) dtdv​=K.v (from the equation (i)) dtdv​=K.(Kx)=K2x a=K2x and v=−Kx+v0​ dtdv​=−K.v=−K(−Kx+v0​) a=K2x−Kv0​ .......(ii) So, according to the equation (ii).

Q. Velocity $(v)$ versus displacement $(x)$ plot of a body moving along a straight line is as shown in the graph. The corresponding plot of acceleration $(a)$ as a function of displacement $(x)$ is

The given graph
$v = K x$ .....(i)

$\frac{d v}{d t} = K . v$ (from the equation (i))
$\frac{d v}{d t} = K . (K x) = K^{2} x$
$a = K^{2} x$
and $v = - K x + v_{0}$
$\frac{d v}{d t} = - K . v = - K (- K x + v_{0})$
$a = K^{2} x - K v_{0}$ .......(ii)
So, according to the equation (ii).