Question

For a particle under going rectilinear motion with uniform acceleration, the magnitude of displacement is one third the distance covered in some time interval. The magnitude of final velocity is less than magnitude of initial velocity for this time interval. Then the ratio of initial speed velocity to the final speed for this time interval is

• A. 1
• B. 2
• C. 3
• D. √2

Answer 1

Answer: (D)

√2

Answer 2

The problem implies a change in direction as displacement is less than distance. This means initial and final velocities have opposite signs. Given final speed is less than initial speed, the particle starts, moves forward, stops, and then moves backward, but does not cross its starting point with a speed greater than its initial speed.

Let initial speed be $|u|$ and final speed be $|v|$. Let acceleration be $-|a|$.

Displacement $s = \frac{|u|^2 - |v|^2}{2|a|}$.

Distance $d = \frac{|u|^2 + |v|^2}{2|a|}$.

Given $|s| = \frac{1}{3} d$. Since $|u|>|v|$ is required for $s>0$, we use $s = \frac{|u|^2 - |v|^2}{2|a|}$.

Substitute these into the given relation: $\frac{|u|^2 - |v|^2}{2|a|} = \frac{1}{3} \left( \frac{|u|^2 + |v|^2}{2|a|} \right)$.

Simplify: $3(|u|^2 - |v|^2) = |u|^2 + |v|^2$.

$3|u|^2 - 3|v|^2 = |u|^2 + |v|^2$.

$2|u|^2 = 4|v|^2$.

$|u|^2 = 2|v|^2$.

The ratio of initial speed to final speed is $\frac{|u|}{|v|} = \sqrt{2}$.