Question

The acceleration of a particle which moves along the positive x-axis varies with its position as shown. If the velocity of the particle is $0.8m/s$ at $x=0$, the velocity of the particle at $x=1.4$ is (in m/s).

• A. $1.6$
• B. $1.2$
• C. $1.4$
• D. None of these

Answer 1

The correct option is D None of these

Velocity at $x=1.4$ is $V$

$a=V\frac{dv}{dx}\int adx=\int vdv$

(from under curve from $0tox=1.4$) $=\int vdv$

$0.4\times 0.4+(\frac{1}{2}\times 0.2\times 0.4+0.4\times 0.2)+0.6\times 0.2=\frac{V^2}{2}=0.16+0.04+0.08+0.12=\frac{V^2}{2}=0.4=\frac{V^2}{2}{V}^2=0.8V=\sqrt{0.8}=0.894m/s$