Acceleration (a) -displacement( s ) graph of a particle moving in a straight line is shown in adjacent figure. The initial velocity of the particle is zero. The v-s graph of the particle would be

Question

Acceleration (a) -displacement( s ) graph of a particle moving in a straight line is shown in adjacent figure. The initial velocity of the particle is zero. The v-s graph of the particle would be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Acceleration, a=(d v/d t)=(d v/d s) × (d s/d t)=v (d v/d s) From given graph a=k s So ∫ limits0v v d v=k ∫ limits0s s d s or (v2/2)=(1/2) k s2 or v=√k s i.e. v propto s So v-s graph is a straight line.

Q. Acceleration $(a)$ -displacement( $s$ ) graph of a particle moving in a straight line is shown in adjacent figure. The initial velocity of the particle is zero. The $v - s$ graph of the particle would be

Acceleration, $a = \frac{d v}{d t} = \frac{d v}{d s} \times \frac{d s}{d t} = v \frac{d v}{d s}$
From given graph $a = k s$
So $0 \int v v d v = k 0 \int s s d s$
or $\frac{v ^{2}}{2} = \frac{1}{2} k s^{2}$
or $v = k s$ i.e. $v \propto s$
So $v - s$ graph is a straight line.

Q. Acceleration (a) -displacement( s ) graph of a particle moving in a straight line is shown in adjacent figure. The initial velocity of the particle is zero. The v−s graph of the particle would be

Acceleration, a=dtdv​=dsdv​×dtds​=vdsdv​ From given graph a=ks So 0∫v​vdv=k0∫s​sds or 2v2​=21​ks2 or v=k​s i.e. v∝s So v−s graph is a straight line.

Q. Acceleration $(a)$ -displacement( $s$ ) graph of a particle moving in a straight line is shown in adjacent figure. The initial velocity of the particle is zero. The $v - s$ graph of the particle would be

Acceleration, $a = \frac{d v}{d t} = \frac{d v}{d s} \times \frac{d s}{d t} = v \frac{d v}{d s}$
From given graph $a = k s$
So $0 \int v v d v = k 0 \int s s d s$
or $\frac{v ^{2}}{2} = \frac{1}{2} k s^{2}$
or $v = k s$ i.e. $v \propto s$
So $v - s$ graph is a straight line.