A body starts from rest with uniform acceleration. Its velocity after 2n seconds is V₀ the displacement of body in last ‘n’ seconds is 3V₀n/β, value of β =
Physics · Mixed Chapter
JEE Mains

Question

A body starts from rest with uniform acceleration. Its velocity after 2n seconds is V₀ the displacement of body in last ‘n’ seconds is 3V₀n/β, value of β =

G1 Physics · Accepted Answer

Step 1: Motion from rest with uniform acceleration a gives V₀=a(2n), so a=V₀/(2n). Step 2: Distance during the last n seconds (from t=n to t=2n) is Δs=½a[(2n)²−n²]=3an²/2. Step 3: Substitute a: Δs=3V₀n/4. Step 4: Match Δs=3V₀n/β, obtaining β=4. Answer: 4.