Practicing Success
If a particle is moving such that the velocity acquired is proportional to the square root of the distance covered, then its acceleration is |
a constant $\propto s^2$ $\propto \frac{1}{s^2}$ $\propto s$ |
a constant |
Let v be the velocity of the particle when the distance covered is s. Then, $v \propto \sqrt{s}$ [Given] $\Rightarrow v=\lambda \sqrt{s}$ $\Rightarrow \frac{d v}{d s}=\frac{\lambda}{2 \sqrt{s}}$ $\Rightarrow v \frac{d v}{d s}=\frac{\lambda v}{2 \sqrt{s}}=\frac{\lambda^2}{2}$ = constant. Hence, the acceleration is constant. |