Question

A particle is moving along a straight line along the positive x-axis such that its speed is inversely proportional the the distance from origin $[V\alpha \frac{1}{x}\Rightarrow v=\frac{k}{x}whereKistheproportionalityconstant]$ At r = 0 the particle is at point $A\left(x_{0,}0\right)$ and moving along the positive x-axis with speed $V_{0}m/s$ The graph of motion of the particle for 1/V versus x ( distance from origin) A to B

• A. The time interval of motion from point A to point B is 12.50 sec
• B. The time interval of motion from point A to point B is 18.75 sec
• C. The proportionally constant k is $10m^2/s$
• D. The proportionally constant k is $20m^2/s$

Answer 1

Answer: (B), (C)

The time interval of motion from point A to point B is 18.75 sec
The proportionally constant k is $10m^2/s$

Given : $v=\frac{k}{x}$

From the graph, $v{|}_A=2$ m/s

$\therefore$ $2=\frac{k}{5}$ $⟹k=10m^2/s$

Also $\frac{dx}{dt}=\frac{k}{x}$ $⟹$ ${\int }_5^{20}xdx={\int }_0^{t}kdt$ where $k=10m^2/s$

$\therefore$ $\frac{x^2}{2}{|}_5^{20}=10\times t{|}_0^t$

$\therefore$ $\frac{{20}^2-5^2}{2}=\left(10\right)t$ $⟹t=18.75$ s