A particle moving with constant non-zero acceleration passes through the same position at times t₁ and t₂. The total distance travelled between these instants is D. If Δt=t₂-t₁, then the magnitude of acceleration and speed at either of the two passages are respectively:
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Question

A particle moving with constant non-zero acceleration passes through the same position at times t₁ and t₂. The total distance travelled between these instants is D. If Δt=t₂-t₁, then the magnitude of acceleration and speed at either of the two passages are respectively:

Options

G1 Physics · Accepted Answer

For constant acceleration, the turning point lies midway in time: turning time = (t₁+t₂)/2. Thus the time from either passage to the turning point is h = Δt/2. Let acceleration magnitude be a. Distance from one passage to the turning point is (a/2)h² = (a/2)(Δt²/4) = aΔt²/8. The motion is symmetric, so total distance D = 2(aΔt²/8) = aΔt²/4. Therefore a = 4D/(Δt)². Speed at either passage: v = ah = [4D/(Δt)²](Δt/2) = 2D/Δt. Hence option B is correct.

A particle moving with constant non-zero acceleration passes through the same position at times t₁ and t₂. The total distance travelled between these instants is D. If Δt=t₂−t₁, then the magnitude of acceleration and speed at either of the two passages are respectively: