The displacement and the increase in the velocity of a moving particle in the time interval of t to (t+1) s are 125 m and 50 m / s, respectively. The distance travelled by the particle in (t+2)ᵗʰ s is _________ m.

Solution

Physics sir · 4 steps

With constant acceleration, each next 1-second displacement grows by exactly a × 1², so just add 50.

1. Find the acceleration. The velocity rises by 50 m s⁻¹ in 1 s, so a = 50 m s⁻², constant throughout.
2. Name the intervals. The first interval runs from t to t+1. The (t+2)ᵗʰ second is the very next one, from t+1 to t+2. These are two back-to-back 1-second intervals.

   sₙₑₓₜ - sₚᵣₑᵥ = a (Δ t)² = 50 × 1² = 50 m

3. Why this works. Displacement in a 1-second interval equals the velocity at its midpoint. The midpoints of the two intervals are 1 s apart, so the second midpoint velocity is 125 + 50 = 175 m s⁻¹. The velocity stays positive, so distance equals displacement.
4. So. Distance in the (t+2)ᵗʰ second = 125 + 50 = 175 m.

The common slip: 225. Treating the (t+2)ᵗʰ second as the interval t+2 to t+3 adds 50 twice and gives 225. That interval is the (t+3)ᵗʰ second.

The idea to keep. Under constant acceleration, displacements in successive equal intervals T form an arithmetic progression with common difference aT². Use it whenever consecutive seconds are compared.

Answer from the NTA final answer key. Solution by PrepWiser.